Question

Let $z$ be a function defined as $z_n=\frac{1}{2iy}[\frac{(-1{)}^{n}n!}{(x-iy{)}^{n+1}}-\frac{(-1{)}^{n}n!}{(x+iy{)}^{n+1}}]$, where $i=\sqrt{-1},n\in N$ and $x,y\in R-\left\{0\right\}$. If $z_n$ be the function of $\theta$, where $\theta ={\tan }^{-1}\frac{y}{x}$, then which of the following is/are correct

• A. $z_6\left(\frac{\pi }{2}\right)=-\frac{45}{16}$ when $y=2$
• B. $z_6\left(\frac{\pi }{2}\right)=\frac{45}{16}$ when $y=2$
• C. $\underset{\theta \to 0}{lim}\frac{y}{\theta }\frac{z_5}{z_4}=-6$
• D. $z_4,z_5,z_6$ are in G.P.

Answer 1

Answer: (A), (C)

$\underset{\theta \to 0}{lim}\frac{y}{\theta }\frac{z_5}{z_4}=-6$

$z_n=\frac{1}{2iy}[\frac{(-1{)}^{n}n!}{(x-iy{)}^{n+1}}-\frac{(-1{)}^{n}n!}{(x+iy{)}^{n+1}}]$

Putting $x=r\cos \theta ,y=r\sin \theta$
Where $r=\sqrt{x^2+y^2},\theta ={\tan }^{-1}\left(y/x\right)$

$x-iy=r(\cos \theta -i\sin \theta )=r{e}^{-i\theta }x+iy=r{e}^{i\theta }$

$z_n=\frac{(-1{)}^{n}n!}{2ir\sin \theta }[\frac{1}{r^{n+1}{e}^{-i(n+1)\theta }}-\frac{1}{r^{n+1}{e}^{i(n+1)\theta }}]$
$\Rightarrow z_n=\frac{(-1{)}^{n}n!}{2i\sin \theta r^{n+2}}[e^{i(n+1)\theta }-e^{-i(n+1)\theta }]\Rightarrow z_n=\frac{(-1{)}^{n}n!\sin (n+1)\theta }{\sin \theta r^{n+2}}$

When $\theta =\frac{\pi }{2},y=2\Rightarrow r=2,n=6$
$z_6\left(\frac{\pi }{2}\right)=\frac{6!\times (-1)}{2^8}=-\frac{45}{16}$

$\frac{z_5}{z_4}=-\frac{5}{r}\times \frac{\sin 6\theta }{\sin 5\theta }$

$\underset{\theta \to 0}{lim}\frac{y}{\theta }\frac{z_5}{z_4}=\frac{r\sin \theta }{\theta }(-\frac{5}{r}\times \frac{\sin 6\theta }{\sin 5\theta })=-6$