Question

Let $A\left(z_1\right)$ be the point of intersection of curves $\arg \left(z-2+i\right)=\frac{3\pi }{4}$ and $\arg \left(z+i\sqrt{3}\right)=\frac{\pi }{3}$. $B\left(z_2\right)$ be the point on the curve $\arg \left(z+i\sqrt{3}\right)=\frac{\pi }{3}$ such that $\left|z_2-5\right|$ is minimum and $C\left(z_3\right)$ be the centre of circle $\left|z-5\right|=3$.
[Note : $i^2=-1$]
If $\left|z-z_1\right|=1$ and $\omega =\mathrm{Re}\left(z+2\right)$, then $\omega$ lie on :

• A. real axis
• B. line not passing through the origin
• C. line segment joining (2,0) and (4,0)
• D. circle centred at (1,0) and radius 2.

Answer 1

The correct option is A real axis

$arg(z-2+i)=\frac{3\pi }{4}$

Let $z=x+iy$

$arg(x-2+iy+i)=\frac{3\pi }{4}$

$\frac{1+y}{x-2}=\tan \left(\frac{3\pi }{4}\right)=(-1)$

$x+y=1\to \left(1\right)$

$arg(z+i\sqrt{3})=\frac{\pi }{3}\Rightarrow arg(x+iy+i\sqrt{3})=\frac{\pi }{3}$

$\frac{y+\sqrt{3}}{x}=\tan \frac{\pi }{3}\Rightarrow y=x\sqrt{3}-\sqrt{3}\to \left(2\right)$

Solving $1$ and $2$, we get point to be $A(1,0)$

Therefore, $z_1=1$

$|z-z_1|=1\Rightarrow |z-1|=1$

$\omega =Re(z+2)=Re(x+2+iy)=x+2$

$\forall xϵR$ as $x\ne$imaginary

$\omega$ lies on Real axis