Question

If $\sin \theta =\left(\frac{z-1}{i}\right),$ where $z$ is non real, $\theta$ represents angle of a triangle, then

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

Answer 1

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

If $\theta$ represents angle of a triangle, then $\sin \theta$ must be real.
$\Rightarrow \frac{z-1}{i}$ will be real

Let $z=x+iy,$ then
$\frac{z-1}{i}=\frac{(x+iy)-1}{i}$
$=\frac{(x-1)+iy}{i}=\frac{(x-1)i}{i\times i}+y$
$=y-i(x-1)$

For $y-i(x-1)$ to be real, $x$ must be equal to $1$
then $\sin \theta =y$ exists if $-1\le y\le 1$

$\therefore Re\left(z\right)=1$ and $-1\le Im\left(z\right)\le 1$