Question

Let $C_1$ and $C_2$ be the two curves on the complex plane defined as
$C_1:z+\overline{z}=2|z-1|$
$C_2:arg(z+1+i)=\alpha$
Where \alpha belongs to the interval $(0,\frac{\pi }{2})$ such that curves $C_1$ and $C_2$ have exactly one point in common and which is denoted by $P\left(z_0\right)$
The value of $|z_0|$ is

• A. $2$
• B. $4$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is C

$\sqrt{2}$

$C_1:z+\overline{z}=2|z-1|2x=2|x-1+iy|x^2=(x-1{)}^2+y^2\Rightarrow y^2=2x-1\Rightarrow y^2=2(x-\frac{1}{2})C_2:arg(z+1+i)=\alpha$
curve $C_2$ is a ray emanating from (-1, -1) and making an angle $\alpha$ from the positive real axis $C_1$ and $C_2$ have exactly one common point
$\therefore C_2$ must be a tangent to $C_1$
Solving $C_1$ and $C_2$
$y^2=2(\frac{y+1}{m}-1)-1m{y}^2=2(y+1-m)-mm{y}^2-2y+3m-2=0\Rightarrow D=0\Rightarrow 4-4m(3m-2)=03m^2-2m-1=0\Rightarrow (3m+1)(m-1)=0\Rightarrow m=-\frac{1}{3},1$
$m=-\frac{1}{3}$ rejected
Putting y = x in the curve $C_1$
$x^2=2x-1\Rightarrow (x-1{)}^2=0\Rightarrow x=1\Rightarrow p\equiv (1,1)$
Complex number corresponding to P is
$z_0=1+i|z_0|=\sqrt{2}$