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Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(5)\$, \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0 \end{alignat*} \end{cases}\space\Longleftrightarrow\space\begin{cases} \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\sqrt{2-\sqrt{3}} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}+2103684\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}

Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(5)\$, \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0 \end{alignat*} \end{cases}\space\Longleftrightarrow\space\begin{cases} \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\sqrt{2-\sqrt{3}} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(2103684-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}

Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)} \end{alignat*} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}+2103684\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}

Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(5)\$, \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0 \end{alignat*} \end{cases}\space\Longleftrightarrow\space\begin{cases} \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\sqrt{2-\sqrt{3}} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}+2103684\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}

Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)} \end{alignat*} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}

Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

• $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
• $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.

Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)} \end{alignat*} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}+2103684\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}