Question

If $\sin \theta =\left(\frac{z-1}{i}\right),$ where $z=x+iy,$ $\theta$ represents an angle of a triangle, then

• A. $Re\left(z\right)=0$
• 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. $Re\left(z\right)=1,0<Im\left(z\right)\le 1$

Answer 1

The correct option is D $Re\left(z\right)=1,0<Im\left(z\right)\le 1$

If $\theta$ represents angle of a triangle, then $\sin \theta$ must be real and $\sin \theta \in (0,1]$

Now, $\frac{z-1}{i}$ should be real, 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,
$Im[y-i(x-1\left)\right]=0\Rightarrow x-1=0\Rightarrow x=1$

Also, $\sin \theta \in (0,1]$ and $\sin \theta =y$
$\Rightarrow y\in (0,1]$
Hence, $Re\left(z\right)=1$ and $0<Im\left(z\right)\le 1$