# Division of two complex numbers

Let we have two complex numbers $z_1=r(\cos(k-\theta)+i\sin(k-\theta))$ and $z_2=r(\cos(k-\theta)-i\sin(k-\theta))$ where $k$ is some constant like $\pi/2$. Since $\cos(k-\theta)=\cos(\theta-k)$, therefore $z_2=r(\cos(\theta-k)-i\sin(k-\theta))$. If we find $z_1/z_2$ using the former representation of $z_2$ we get:

$z_1/z_2=1+i\sin2(k-\theta)$ and if we use the latter representation $z_2=r(\cos(\theta-k)+i\sin(k-\theta))$ then:

$z_1/z_2=\cos2(k-\theta)+i\sin2(k-\theta)$

These two are different complex numbers for the same $z_1$ and $z_2$. Explain which one is correct.

• At least one conclusion appears to be predicated on the (incorrect) belief that $$(\cos A + i\sin B)/(\cos C + i\sin D) = \cos(A - C) + i\sin(B - D)$$for all real $A$, $B$, $C$, and $D$. What's true is:$$(\cos A + i\sin A)/(\cos C + i\sin C) = \cos(A - C) + i\sin(A - C)$$for all real $A$ and $C$. – Andrew D. Hwang Sep 4 '16 at 18:51

## 2 Answers

Using the classic method you can:

1) Simplify $r$

2) Multiply both $z_1$ and $z_2$ by $z_2$' conjugate equal to $cos(k-\theta)+isin(k-\theta)$ obtaining:

$$\frac{cos^2(k-\theta)-sin^2(k-\theta)+ i 2cos(k-\theta)sin(k-\theta)}{cos^2(k-\theta)+sin^2(k-\theta)}$$

3) You can notice that the denominator is equal to 1 but I don't see any further useful simplification.

So your second attempt is the right one: $z_1/z_2=\cos(2(k-\theta))+i\sin(2(k-\theta))$

Note that $z_2=\overline{z_1}$ and $z_1=re^{i(k-\theta)}$, so

$$\frac{z_1}{z_2}=\frac{e^{i(k-\theta)}}{e^{-i(k-\theta)}}=e^{2i(k-\theta)}=\cos(2(k-\theta))+i\sin(2(k-\theta)).$$