Question

Match the following:

1. $(i{)}^i$	a) $i\frac{\pi }{2}$
2. log,i	b) $i\frac{\pi }{2}+\log \frac{\pi }{2}$
3. log (log i)	c)$\sqrt{2}$
4. $\sqrt{i}+\sqrt{-i}$	d) $e^{-\frac{\pi }{2}}$
	e) $e^{\frac{\pi }{2}}$

• A. $1-a,2-b,3-c,4-d$
• B. $1-d,2-a,3-b,4-c$
• C. $1-b,2-a,3-c,4-d$
• D. $1-b,2-b,3-a,4-d$

Answer 1

The correct option is B $1-d,2-a,3-b,4-c$

(1) $i=e^{i^{\pi }{/}_{2}i}=e^{{}^{-}\pi {/}_2}=option\left(d\right)$
(2) $logi=lo{g}_e\left|1\right|e^{i^{\pi }{/}_2}$
$=0+i^{\pi }{/}_2=option\left(a\right)$
(3) $log\left(logi\right)=log\left(i^{\pi }{/}_2\right)aslogi=i^{\pi }{/}_2$
$=log\left(\frac{\pi }{2}{e}^{i{}^{\{}pi}{/}_2\right)$
$=log\frac{\pi }{2}+lo{g}_e{e}^{i^{\pi }{/}_2}$
$=log\frac{\pi }{2}+i\frac{\pi }{2}=option\left(b\right)$
(4) $\sqrt{i}+\sqrt{-1}=e^{\left(i^{\pi }{/}_2\right)\times \frac{1}{2}}+e^{(-i^{\pi }{/}_2)\times \frac{1}{2}}$
$=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$
$\sqrt{2}=option\left(c\right)$