Question

If Arg$\left(z+a\right)=\frac{\pi }{6}$ and Arg$\left(z-a\right)=\frac{2\pi }{3};a\in R^{+}$ , then

• A. $z$ is independent of $a$
• B. $\left|a\right|=\left|z+a\right|$
• C. $z=aCis\frac{\pi }{6}$
• D. $z=aCis\frac{\pi }{3}$

Answer 1

Answer: (D)

$z=aCis\frac{\pi }{3}$

We have,

$\arg (z+a)=\frac{\pi }{6}$

$\arg (z-a)=\frac{2\pi }{3}$

$z+a=r_1{e}^{i\frac{\pi }{6}}\Rightarrow z=-a+r_1{e}^{i\frac{\pi }{6}}$ ……. (1)

$z-a=r_2{e}^{i\frac{2\pi }{3}}\Rightarrow z=a+r_2{e}^{i\frac{2\pi }{3}}$ …….. (2)

On equating both equations, we get

$-a+r_1{e}^{i\frac{\pi }{6}}=a+r_2{e}^{i\frac{2\pi }{3}}$

$-a+r_1(\frac{\sqrt{3}}{2}+\frac{1}{2}i)=a+r_2(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

On comparing real part and imaginary part, we get

$\frac{r_1}{2}=\frac{\sqrt{3}}{2}{r}_2\Rightarrow r_1=\sqrt{3}{r}_2$

$-a+\frac{\sqrt{3}}{2}{r}_1=a-\frac{1}{2}{r}_2$

$-a+\frac{\sqrt{3}}{2}\times \sqrt{3}{r}_2=a-\frac{1}{2}{r}_2$

$\frac{3}{2}{r}_2=2a-\frac{1}{2}{r}_2$

$2r_2=2a$

$r_2=a$

$r_1=\sqrt{3}a$

Therefore, the complex number is,

$z=a+a(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

$z=a(\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

$z=aC$ is $\frac{\pi }{3}$

Hence, this is the value of $z$.