Question

If $\arg \left(z\right)=\frac{\pi }{3}$ and $\arg (z-1)=\frac{2\pi }{3},$ then $z$ is

• A. $\frac{1}{2}+i\frac{\sqrt{3}}{2}$
• B. $\sqrt{3}+i$
• C. $1+i\sqrt{3}$
• D. $\frac{\sqrt{3}}{2}+i\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}+i\frac{\sqrt{3}}{2}$

Let $z=x+iy$
Given : $\arg \left(z\right)=\frac{\pi }{3}$
$\Rightarrow \frac{y}{x}=\tan \left(\arg z\right)=\sqrt{3}$
$\Rightarrow y=x\sqrt{3}$

Now, $z-1=(x+iy)-1=(x-1)+iy$
Since $\arg (z-1)=\frac{2\pi }{3}$, therefore $z-1$ lies in $2^{nd}$ quadrant in Argand plane.
$\therefore x-1<0$ and $y>0$
Now, $\pi -\alpha =\arg (z-1)$
$\Rightarrow \alpha =\frac{\pi }{3}$
$\tan \alpha =\left|\frac{y}{x-1}\right|$
$\Rightarrow \sqrt{3}=\left|\frac{y}{x-1}\right|$
$\Rightarrow \frac{y}{1-x}=\sqrt{3}$
$\Rightarrow x\sqrt{3}=\sqrt{3}(1-x)$ $(\because y=x\sqrt{3})$
$\Rightarrow 2x=1$
$\Rightarrow x=\frac{1}{2}$ and $y=\frac{\sqrt{3}}{2}$

Thus, the complex number is $\frac{1}{2}+i\frac{\sqrt{3}}{2}$