Question

${\sin }^{-1}\left\{\frac{1}{i}(z-1)\right\}$, where $z$ is a complex number, can be the angle of a triangle if

• A. $Re\left(z\right)=1,Img\left(z\right)=2$
• B. $Re\left(z\right)=1,0<Img\left(z\right)\le 1$
• C. $Re\left(z\right)+Img\left(z\right)=0$
• D. none of these

Answer 1

Answer: (B)

$Re\left(z\right)=1,0<Img\left(z\right)\le 1$

${\sin }^{-1}\left\{\frac{1}{i}(z-1)\right\}$

Given that $x$ is a complex number,

$\Rightarrow z={\sin }^{-1}\left\{\frac{1}{i}(z-1)\right\}$

In the algebric form : $z=0$

Exponential form, $z=ox{e}^{i\theta }=0$

The polar form $=r=\left|z\right|=0$

$\xi$ $argz=0=0\pi$

$\Rightarrow$ the $Re\left(z\right)=1,$

and $0<Img\left(z\right)\le 1$

Hence, the answer is $Re\left(z\right)=1,0<Img\left(z\right)\le 1.$