Question

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

• A. $Re\left(z\right)=1,I_m\left(z\right)=2$
• B. $Re\left(z\right)=1,-1\le I_m\left(z\right)\le 1$
• C. $Re\left(z\right)+I_m\left(z\right)=0$
• D. $noneofthese$

Answer 1

The correct option is B $Re\left(z\right)=1,-1\le I_m\left(z\right)\le 1$

Let $z=x+iy$

$z-1$$=(x-1)+iy$

$\frac{z-1}{i}$$=i-iz$

$=i(1-z)$
$=-y+i(x-1)$
$=A$

Now for

$si{n}^{-1}\left(A\right)$ A has to be real.

$\therefore x-1=0$
$\Rightarrow x=0$

$Real\left(z\right)=0$.
And
$Aϵ[-1,1]$
$\therefore yϵ[-1,1]$

Hence $Im\left(z\right)ϵ[-1,1]$