Question

If the value of $\frac{(1+i{)}^{1-i}}{(1-i{)}^{1+i}}$ = isin p + cos p. Find the value of p.

• A. 1
• B. 2
• C. 3
• D.

Answer 1

The correct option is B

2

z = $\frac{(1+i{)}^{1-i}}{(1-i{)}^{1+i}}$

Taking ${log}_e$ on both sides

${log}_e$z = (1-i)${log}_e(1+i)-(1+i){log}_e(1-i)$

= (1-i)${log}_e[\sqrt{2}.e^{\frac{i\pi }{4}}]-(1+i){log}_e[\sqrt{2}.e^{i\left(\frac{\pi }{4}\right)}]$

= (1-i) $[{log}_e\sqrt{2}+\frac{i\pi }{4}]$ - (1+i) $[{log}_e\sqrt{2}+(-\frac{i\pi }{4})]$

= ${log}_e\sqrt{2}+\frac{i\pi }{4}-i{log}_e\sqrt{2}-\frac{i^2\pi }{4}-{log}_e\sqrt{2}+\frac{i\pi }{4}-i{log}_e\sqrt{2}+\frac{i^2\pi }{4}$

= $\frac{i\pi }{2}-2i{log}_e\sqrt{2}$

${log}_{e}z=\frac{i\pi }{2}-i{log}_{e}2$

z = $e^{i\frac{\pi }{2}}.e^{i{log}_{e}2}$

= $\left[(cos\frac{\pi }{2}+isin\frac{\pi }{2})\right(cos\left({log}_{e}2\right)+isin\left({log}_{e}2\right)\left)\right]$

= 1 $\times$ [cos(log 2) + i sin(log 2) ]

z = [cos(${log}_{e}2$) + i sin(${log}_{e}2$)]

From the above equation, we can see that p = ${log}_{e}2$