Question

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

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

Answer 1

Answer: (A), (D)

Let
$z=x+iy$
$(z-a)$
$=(x-a)+i\left(y\right)$
Argument is $\frac{2\pi }{3}$
Hence
$\frac{(x-a{)}^2}{(x-a{)}^2+y^2}=\frac{1}{4}$
$3(x-a{)}^2=y^2$ ...(i)
$z+a$
$=(x+a)+iy$
Now argument is $\frac{\pi }{6}$
Hence
$\frac{y^2}{(x+a{)}^2+y^2}=\frac{1}{4}$
$3y^2=(x+a{)}^2$
$9(x-a{)}^2=(x+a{)}^2$
$(x+a)=\pm (3x-3a)$
$x+a=3x-3a$
$x=2a$ and $x+a=-3x+3a$
$x=\frac{a}{2}$
Hence
$3y^2=(x+a{)}^2$
$3y^2=9a^2$
$y=\pm \sqrt{3}a$
$z=a(2\pm \sqrt{3}i)$...(n)
or
$3y^2=\frac{9a^2}{4}$
$y=\pm \frac{\sqrt{3}a}{2}$
$z=a\left(\frac{1\pm \sqrt{3}i}{2}\right)$
Hence
$|z|=|a\left(\frac{1\pm \sqrt{3}i}{2}\right)|=a$
Now we consider $z=a\left(\frac{1+\sqrt{3}i}{2}\right)$
$arg\left(z\right)=\frac{\pi }{3}$
Hence, option 'D' is correct.