Question

Let $z=x+iy$ & amp $\left(e^{z^2}\right)$ = amp $\left(e^{(z+i)}\right)$. If $y=\left(x\right)$ is a function, then $y\left(3\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (D)

$\frac{1}{5}$

$\left(e^{z^2}\right)=e^{(x^2-y^2)+2ixy}=e^{x^2-y^2}.e^{ixy}$
$\Rightarrow$ amp $\left(e^{z^2}\right)=2xy$
And amp$\left(e^{(z+i)}\right)=$ amp$\left(e^{x+(y+1)i}\right)=$ amp$(e^x.e^{i(y+1)})=(y+1)$
$\therefore 2xy=y+1\Rightarrow y=\frac{1}{(2x-1)}$

$\therefore y\left(3\right)=\frac{1}{5}$