1
$\begingroup$

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: $$\frac {\sin(5x)}{\sin(x)}=\frac {\textrm{Im}[(\cos(x)+i\sin(x))^5]}{\textrm{Im}[(\cos(x)+i\sin(x))]}=\textrm{Im}\left[\frac {(\cos(x)+i\sin(x))^5}{(\cos(x)+i\sin(x))}\right]=\textrm{Im}[(\cos(x)+i\sin(x))^4]=\dots$$

$\endgroup$
  • $\begingroup$ No, this identity is just incorrect. $\endgroup$ – Amitai Yuval Jun 10 '15 at 21:02
  • $\begingroup$ The answer is $U_4(\cos x)$, i.e. the Chebyshev polynomial of the second kind $\endgroup$ – uranix Jun 10 '15 at 21:09
1
$\begingroup$

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 $\Im\frac{z_1}{z_2}=\Im 1 = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.