Question

If $e^{x+iy}=\alpha +i\beta ,$ then $x+iy$ is called logarithm of $\alpha +i\beta$ to the base e.$\therefore {\log }_e(x+iy)={\log }_e\left(r{e}^{i\theta }\right)$ $={\log }_{e}r+i\theta$ where r is modulus value of $x+iy$ & $\theta$ be the argument of $x+iy$ If $i^{\alpha +i\beta )}=\alpha +i\beta ,$ then ${\alpha }^2+{\beta }^2$ equals

• A. $e^{(2n+1)\frac{\pi \alpha }{2}}$
• B. $e^{-\pi \beta }$
• C. $e^{(4n+1)\alpha \beta }$
• D. None of these

Answer 1

Answer: (B)

$e^{-\pi \beta }$

$i^{(\alpha +i\beta )}=\alpha +i\beta$
$\Rightarrow (\alpha +i\beta )\log i=\log (\alpha +i\beta )$
$\Rightarrow \log (\alpha +i\beta )=(\alpha +i\beta )[i\frac{\pi }{2}+2n\pi i],nϵI$
$\Rightarrow \frac{1}{2}\log ({\alpha }^2+{\beta }^2)+i{\tan }^{-1}\frac{\beta }{\alpha }=i(\alpha +i\beta )(4n+1)\frac{\pi }{2}$
Equating imaginary parts we have
$\frac{1}{2}\log ({\alpha }^2+{\beta }^2)=-\frac{-\beta \pi }{2}(4n+1)$ $\Rightarrow {\alpha }^2+{\beta }^2=e^{-(4n+1)\pi \beta }$