Question

If P(z) and A(z₁) two be variable points such that $zz_1 = |z|^2$ and $|z-z| + |z_1 + z| = 10$ then area enclosed by the curve formed by them

• A. 25π
• B. 20π
• C. 50
• D. 100

Answer 1

Answer: (C)

50

Answer 2

We are given that for variable points with complex numbers $z$ and $z_1$ the conditions hold:

$$zz_1 = |z|^2 \quad \text{and} \quad |z-\bar{z}|+|z+\bar{z}|=10.$$

Notice that

$$zz_1 = |z|^2 \implies z_1 = \frac{|z|^2}{z}.$$

Writing $z$ in polar form, $z = re^{i\theta}$, we have:

$$\frac{|z|^2}{z} = \frac{r^2}{re^{i\theta}} = re^{-i\theta} = \bar{z}.$$

Thus, $z_1=\bar{z}$.

Let $z=x+iy$ so that:

$$z+\bar{z}=2x \quad \text{and} \quad z-\bar{z}=2iy.$$

Taking moduli,

$$|z-\bar{z}|=2|y| \quad \text{and} \quad |z+\bar{z}|=2|x|.$$

The given condition becomes:

$$2|y|+2|x|=10 \implies |x|+|y|=5.$$

This represents a diamond (a square rotated by 45°) with vertices at $(5,0)$, $(0,5)$, $(-5,0)$, and $(0,-5)$.

The area $A$ of a diamond with diagonals $d_1$ and $d_2$ is

$$A=\frac{d_1\cdot d_2}{2}.$$

Here, both diagonals are $10$, so

$$A=\frac{10\cdot10}{2}=50.$$