Question

Let $C_1$ and $C_2$ be two curves on the complex plane defined as $C_1: z+\overline{z}=2(|z-1|)$ $C_2: arg(z+1+i) = \alpha$ where $\alpha \in (0, \pi)$ such that the curves $C_1$ and $C_2$ touches each other at $P(z_0)$ then the value of $|z_0|^2$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Solution:

Given the curves

$$C_1: z+\overline{z}=2|z-1| \quad\text{and}\quad C_2: \arg(z+1+i)=\alpha, \quad \alpha\in (0,\pi).$$

Let $z=x+iy$. Then:

• For $C_1$:

  $$z+\overline{z}=2x, \quad |z-1|=\sqrt{(x-1)^2+y^2}.$$

  Hence,

  $$2x=2\sqrt{(x-1)^2+y^2}\quad \Longrightarrow\quad x=\sqrt{(x-1)^2+y^2}.$$

  Squaring both sides:

  $$x^2=(x-1)^2+y^2 \quad\Longrightarrow\quad x^2=x^2-2x+1+y^2.$$

  Therefore,

  $$y^2+1-2x=0 \quad\Longrightarrow\quad 2x=y^2+1 \quad\Longrightarrow\quad x=\frac{y^2+1}{2}.$$

  This represents a parabola with equation $y^2=2x-1$.

• For $C_2$:
  Write $z+1+i=(x+1)+(y+1)i$. The condition

  $$\arg(z+1+i)=\alpha$$

  implies that the point $(x,y)$ lies on the straight line

  $$y+1=\tan\alpha\,(x+1),$$

  or equivalently,

  $$y=\tan\alpha\,(x+1)-1.$$

  Let $m=\tan\alpha$; then $C_2$ is the line

  $$y=m(x+1)-1.$$

Since $C_1$ and $C_2$ touch each other at $P(z_0)$, the line is tangent to the parabola $y^2=2x-1$. Substitute $y=m(x+1)-1$ into the parabola:

$$[m(x+1)-1]^2=2x-1.$$

Expanding,

$$m^2(x+1)^2-2m(x+1)+1=2x-1.$$

Express this as a quadratic in $x$:

$$m^2x^2+[2m^2-2m-2]x+(m^2-2m+2)=0.$$

For tangency, the discriminant must be zero:

$$\Delta = [2m^2-2m-2]^2-4m^2\,(m^2-2m+2)=0.$$

Factor out 4:

$$4\Big[(m^2 - m -1)^2-m^2(m^2-2m+2)\Big]=0.$$

Dividing by 4:

$$(m^2 - m -1)^2-m^2(m^2-2m+2)=0.$$

Expanding and simplifying,

$$m^4-2m^3-m^2+2m+1 - (m^4-2m^3+2m^2)=0,$$ $$-3m^2+2m+1=0 \quad\Longrightarrow\quad 3m^2-2m-1=0.$$

Solving,

$$m=\frac{2\pm \sqrt{4+12}}{6}=\frac{2\pm4}{6}.$$

Thus, $m=1$ or $m=-\frac{1}{3}$. Since $\alpha \in (0, \pi)$ gives $\tan \alpha > 0$, we take $m=1$ which implies $\alpha=\frac{\pi}{4}$.

Now, with $m=1$, the line $C_2$ becomes:

$$y=1\cdot (x+1)-1=x.$$

Substitute $y=x$ in the parabola:

$$x^2=2x-1\quad \Longrightarrow\quad x^2-2x+1=0\quad \Longrightarrow\quad (x-1)^2=0.$$

So, $x=1$ and $y=1$. That is,

$$z_0=1+i.$$

Then,

$$|z_0|^2=1^2+1^2=2.$$

Explanation of the minimal solution:

1. Convert $C_1$ to a parabola $y^2=2x-1$.
2. Express $C_2$ as the line $y=\tan\alpha (x+1)-1$.
3. Use tangency condition (discriminant = 0) to find $\tan\alpha=1$.
4. Find the point of tangency $(x,y)=(1,1)$ so that $z_0=1+i$.
5. Therefore, $|z_0|^2=2$.