# Why is the phase angle in the summation of the phasors the negative arctan?

I was reading a textbook and came across the following

$$\\I_1 = Ae^{j0} = A \\I_2 = Be^{-j\frac{\pi}{2}} \\\bf{I} = I_1 + I_2 \\If\ A=10\ and\ B=20 \\|I| = \sqrt{A^2 + B^2} \\\angle I = -tan^{-1}(B/A) \\I = 27.98e^{j30.4^{\circ}}$$

From my knowledge, of complex numbers, shouldn't $$\\angle\bf{I} \$$ be $$\\angle \bf{I} = tan^{-1}(B/A)\$$ since both B and A are in the first quadrant?

• You can't just use the values of A and B since they are magnitudes of the phasor. They will always be positive since they are magnitudes. – DKNguyen Feb 1 at 6:25

I = A - jB is a complex number and it is in the fourth quadrant. So argument of I must be negative. Mind that I_2 has angle -pi/2.

Positive x and negative y means Quadrant IV.

$$\\vec {I_1} = 10\ ∡ 0^\circ\ A\$$

$$\\vec {I_2} = 20\ ∡ - \frac {\pi} {2}\ A = 20\ ∡ - 90^\circ\ A \$$

$$\\vec {I_1} + \vec {I_2} = 10 - j20\ A = 22.36\ ∡ -63.43^\circ\ A \$$

This does not agree with your $$\ 27.98\ e^{j30.4^\circ} \$$, but that is the math.

y is larger than x, so angle has to be greater than $$\45^\circ\$$.