Question

If $z$ is a uni-modular complex number such that $\mathrm{Re}(z-1)+\mathrm{Re}\left(z^2\right)=$ $\underset{0}{\overset{\pi /2}{\int }}\sin x\cdot \ln |\sin x-\cos x|dx$ then

• A. $z+\bar{z}=-2$
• B. $z+\overline{z}=1$
• C. $\arg \cdot (z)=\frac{\pi }{3}$
• D. $\arg .(z)=\pi$

Answer 1

Answer: (A), (B), (C), (D)

$I=\underset{0}{\overset{\pi /2}{\int }}\sin x\cdot \ln |\sin x-\cos x|dx$
$I=\underset{0}{\overset{\pi /2}{\int }}\cos x\cdot \ln |\sin x-\cos x|dx$
$\therefore 2I=\underset{0}{\overset{\pi /2}{\int }}(\sin x+\cos x)\ln |\sin x-\cos x|dx$
$=\underset{0}{\overset{\pi /4}{\int }}(\sin x+\cos x)\ln (\cos x-\sin x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}(\sin x+\cos x)\ln (\sin x-\cos x)dx$
$=[(s-c)\ln (c-s){]}_0^{\pi /4}-\underset{0}{\overset{\pi /4}{\int }}\frac{s-c}{c-s}(-s-c)dx+[(s-c)\ln (s-c){]}_{\pi /4}^{\pi /2}-\underset{\pi /4}{\overset{\pi /2}{\int }}(s-c)\cdot \frac{1}{(s-c)}(c+s)dx$
$=-2$
$\therefore I=-1$
$\therefore \mathrm{Re}(z-1)+\mathrm{Re}{z}^2=-1$
$\text{ Let }z=e^{i\theta }$
$\therefore \cos 2\theta +\cos \theta =0\Rightarrow \cos \theta =-1,\cos \theta =\frac{1}{2}$
$\text{ when }\cos \theta =-1\Rightarrow z=-1$
$\text{ when }\cos \theta =\frac{1}{2}\Rightarrow z=\frac{1}{2}+\frac{i\sqrt{3}}{2}$