Question

I. The value of $i^i$ is $e^{\frac{-\pi }{2}}$

II. $z=ilog(2+\sqrt{3})$ then $sinz=2$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. Neither I nor II are true

Answer 1

The correct option is A only I is true

$i^i$$=(e^{i\frac{\pi }{2}}{)}^i$
$=e^{\frac{-\pi }{2}}$
$z=ilog(2+\sqrt{3})$
$sin\left(z\right)=\frac{e^{iz}-e^{iz}}{2}$
And
$ilog(2+\sqrt{3})$
$=ilog\left(\frac{1}{2-\sqrt{3}}\right)$
$iz=-log\left(\frac{1}{2-\sqrt{3}}\right)$
$=log(2-\sqrt{3})$
Hence
$sin\left(z\right)=\frac{e^{iz}-e^{-iz}}{2}$
$=\frac{e^{2iz}-1}{2e^{iz}}$
$=\frac{(2-\sqrt{3}{)}^2-1}{2(2-\sqrt{3})}$
$=\frac{4+3-2\sqrt{3}-1}{2(2-\sqrt{3})}$
$=\frac{6-2\sqrt{3}}{2(2-\sqrt{3})}$
$=(3-\sqrt{3})(2+\sqrt{3})$
$=6+3\sqrt{3}-2\sqrt{3}-3$
$=3+\sqrt{3}$
Hence, option 'A' is correct.