# Help understanding AC circuit voltage division math in frequency domain

I don't fully understand the section b. How does the math in fhat voltage division equation work out to be 200 ang(-113.13) volts? The voltage division equation is $$V=\frac{-j40}{90+j160-j40}(750\angle30)$$

You can combine $$j160-j40$$ in the denominator to give $$j120$$ and convert $$750\angle30$$ from polar to complex rectangular $$(x,\ jy)$$ form by: $$(750\cos(30),\ j750\sin(30))$$. Now we have:

$$V=\frac{-j40\times(649.5+j375)}{90+j120}$$ $$=\frac{-j25980+15000}{90+j120}$$

Convert back to phasor notation $$A\angle B: A=\sqrt{x^2+y^2},\ \ B=\tan^{-1} (\frac{y}{x})$$ which gives:

$$V = \frac{30000\angle -60}{150\angle 53.13}$$

In phasor notation this simplifies to:

$$V=30000/150 \angle (-60-53.13) = 200\angle-113.13$$

This equation is using complex arithmetic, where each value has both a real and an imaginary component. Complex numbers can be expressed either in rectangular form $$R + jI$$ or in polar form $$M\angle\theta$$. There are many online resources that will help you learn to use complex arithmetic.