Question

If $z$ is a complex number satisfying $\arg (z+a)=\frac{\pi }{6}$ and $\arg (z-a)=\frac{2\pi }{3},a\in R^{+}$, then

• A. $|z|=a$
• B. $|z|=2a$
• C. $\arg \left(z\right)=\frac{\pi }{2}$
• D. $\arg \left(z\right)=\frac{\pi }{3}$

Answer 1

Answer: (A), (D)

$\arg \left(z\right)=\frac{\pi }{3}$

Given : $\arg (z+a)=\frac{\pi }{6}$ and $\arg (z-a)=\frac{2\pi }{3}$

So, $z$ lies on the circle whose diameter is $AB$
$OB=a=OC$
Therefore,
$\angle COB=\frac{\pi }{3}\Rightarrow z=a(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$

Hence,
$|z|=a\arg \left(z\right)=\frac{\pi }{3}$