# Expressing $\frac {\sin(5x)}{\sin(x)}$ in powers of $\cos(x)$ using complex numbers

Am I correct in thinking that if $z_1=a+ib$ and $z_2=c+id$, then it is not generally true that $$\frac {\textrm{Im}(z_1)}{\textrm{Im}(z_2)} = \textrm{Im}\left(\frac {z_1}{z_2}\right)$$ I did a division of the two and got $\textrm{Im}(\frac {z_1}{z_2})=\frac{bc-ad}{c^2+d^2}$ whereas $\frac {\textrm{Im}(z_1)}{\textrm{Im}(z_2)}=\frac bd$.

I'm trying to express $\frac {\sin(5x)}{\sin(x)}$ in powers of $\cos(x)$:

$$\frac {\sin(5x)}{\sin(x)}=\frac {\textrm{Im}[(\cos(x)+i\sin(x))^5]}{\sin(x)}=\dots$$ and I know how to go that way, but I want to know if I can do this: $$\frac {\sin(5x)}{\sin(x)}=\frac {\textrm{Im}[(\cos(x)+i\sin(x))^5]}{\textrm{Im}[(\cos(x)+i\sin(x))]}=\textrm{Im}\left[\frac {(\cos(x)+i\sin(x))^5}{(\cos(x)+i\sin(x))}\right]=\textrm{Im}[(\cos(x)+i\sin(x))^4]=\dots$$

• No, this identity is just incorrect. – Amitai Yuval Jun 10 '15 at 21:02
• The answer is $U_4(\cos x)$, i.e. the Chebyshev polynomial of the second kind – uranix Jun 10 '15 at 21:09

You are correct to think that this is not generally true. For example, if $z_1=z_2=i$, then $\frac{\Im{z_1}}{\Im{z_2}}=\frac{1}{1}=1$, but $\Im\frac{z_1}{z_2}=\Im 1 = 0$.