Question

If $z$ is a uni-modular complex number such that $\operatorname{Re}( z -1)+\operatorname{Re}\left( z ^2\right)=$ $\int\limits_0^{\pi / 2} \sin x \cdot \ln |\sin x-\cos x| d x$ 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)

$I=\int\limits_0^{\pi / 2} \sin x \cdot \ln |\sin x-\cos x| d x$
$I =\int\limits_0^{\pi / 2} \cos x \cdot \ln |\sin x -\cos x | dx $
$\therefore 2 I =\int\limits_0^{\pi / 2}(\sin x +\cos x ) \ln |\sin x -\cos x | dx$
$=\int\limits_0^{\pi / 4}(\sin x +\cos x ) \ln (\cos x -\sin x ) dx +\int\limits_{\pi / 4}^{\pi / 2}(\sin x +\cos x ) \ln (\sin x -\cos x ) dx $
$=[( s - c ) \ln ( c - s )]_0^{\pi / 4}-\int\limits_0^{\pi / 4} \frac{ s - c }{ c - s }(- s - c ) dx +[( s - c ) \ln ( s - c )]_{\pi / 4}^{\pi / 2}-\int\limits_{\pi / 4}^{\pi / 2}( s - c ) \cdot \frac{1}{( s - c )}( c + s ) dx$
$=-2$
$\therefore I =-1 $
$\therefore \operatorname{Re}( z -1)+\operatorname{Re} z ^2=-1 $
$\text { Let } z = e ^{ i \theta}$
$\therefore \cos 2 \theta+\cos \theta=0 \Rightarrow \cos \theta=-1, \cos \theta=\frac{1}{2}$
$\text { when } \cos \theta=-1 \Rightarrow z =-1$
$\text { when } \cos \theta=\frac{1}{2} \Rightarrow z =\frac{1}{2}+\frac{ i \sqrt{3}}{2}$