I am having some trouble understanding this type of transformation. The materials provided by my professor doesn't even mention the method that is being used to switch from complex to sinusoidal and vise versa. For instance

I= -10(1 +j sqrt(3)/3)) .


i= 20/3 sqrt(6) sin(wt +210).

I only understand that 180 degrees are added because of the minus in the first part of the equation but what about the rest ?

here is another example in the opposite matter i(t) = 6 sin (wt + 3pi/4).


I= -3+3j


The basic idea is that to any sinusoidal quantity \$x(t)\$ with angular frequency \$\omega\$ you can associate a constant complex quantity \$X\$ so that $$x(t) = \operatorname{Re}[X\mathrm{e}^{\mathrm{j}\omega t}],\qquad\qquad(1)$$ where \$\operatorname{Re}\$ denotes the real part of the complex number \$X\mathrm{e}^{\mathrm{j}\omega t}\$. The quantity \$X\$ is called the phasor associated to \$x(t)\$. This kind of representation is also called Steinmentz transform.

Other representations are possible, e.g, the associated phasor can be chosen so that $$x(t) = \operatorname{Re}[\sqrt{2}\,X\mathrm{e}^{\mathrm{j}\omega t}],\qquad\qquad(2)$$ in this way the modulus \$|X|\$ of the complex number \$X\$ yields the root-mean-square (RMS) value of \$x(t)\$. Or one can choose the phasor so that \$x(t)\$ is represented by the imaginary parts $$x(t) = \operatorname{Im}[X\mathrm{e}^{\mathrm{j}\omega t}],\qquad\qquad(3)$$ or $$x(t) = \operatorname{Im}[\sqrt{2}\,X\mathrm{e}^{\mathrm{j}\omega t}].\qquad\qquad(4)$$

In the case of your examples, with a bit of "reverse engineering", it's easy to see that your professor chose the representation (4).

Let's consider your last example with \$I = (-3+3\mathrm{j})\,\mathrm{A}\$ (complex quantities have units too!). The polar representation of \$I\$ is $$I = (-3+3\mathrm{j})\,\mathrm{A} = 3\sqrt{2}\mathrm{e}^{\mathrm{j}\frac{3\pi}{4}}\,\mathrm{A}.$$ Thus, $$i(t) = \operatorname{Im}[\sqrt{2}\,I\mathrm{e}^{\mathrm{j}\omega t}] = \operatorname{Im}\left[6\mathrm{e}^{\mathrm{j}\left(\omega t+\frac{3\pi}{4}\right)}\,\mathrm{A}\right],$$ and by using the Euler's formula $$\mathrm{e}^{\mathrm{j}\alpha} = \cos\alpha+\mathrm{j}\sin\alpha,$$ we obtain $$\begin{align}i(t) &= \operatorname{Im}\left\{6\left[\cos\left(\omega t+\frac{3\pi}{4}\right)+\mathrm{j}\sin\left(\omega t+\frac{3\pi}{4}\right)\right]\,\mathrm{A}\right\}, \\ &= 6\sin\left(\omega t+\frac{3\pi}{4}\right)\,\mathrm{A}.\end{align}$$


The "big I" notation is called "phasor notation", which is a shorthand notation that only represents the phase and magnitude of a signal, while ignoring its frequency. It uses rectangular coordinates, with a real component and an imaginary component. The general form is

$$I = Real + j\cdot Imaginary$$

The "i(t)" notation is a direct representation of the same signal in the time domain, but now the phase and magnitude are given using polar coordinates. The general form is

$$i(t) = A \sin(\omega t + \theta)$$

where A is the amplitude, ω is the radian frequency and θ is the phase angle.

You can look up "rectangular to polar" and "polar to rectangular" coordinate conversions to convert between the two notations.


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