Question

Let \(z\) and \(w\) be two distinct non-zero complex numbers. If \[ |z|^2w-|w|^2z=z-w, \] then

• A. \(w=\overline{z}^{\,2}\)
• B. \(zw=2\)
• C. \(z\overline{w}=1\)
• D. \(w=\overline{z}\)

Hint:

For complex numbers, use \(|z|^2=z\overline{z}\). This often helps in converting modulus equations into equations involving conjugates.

Answer 1

The Correct Option is C

Step 1: Use modulus form.
Given, \[ |z|^2w-|w|^2z=z-w \] We know that \[ |z|^2=z\overline{z} \] and \[ |w|^2=w\overline{w} \] So, \[ z\overline{z}w-w\overline{w}z=z-w \] Taking \(zw\) common from the left side, \[ zw(\overline{z}-\overline{w})=z-w \]

Step 2: Express using \(z-w\).
Since \[ \overline{z}-\overline{w}=\overline{z-w}, \] we get \[ zw\overline{(z-w)}=z-w \] As \(z\) and \(w\) are distinct, \[ z-w\neq 0 \] Therefore, \[ zw=\frac{z-w}{\overline{z-w}} \]

Step 3: Write \(z-w\) in polar form.
Let \[ z-w=re^{i\theta}, \] where \[ r\gt 0 \] Then, \[ \overline{z-w}=re^{-i\theta} \] Hence, \[ zw=\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta} \]

Step 4: Express \(z\) and \(w\) in a common direction.
Since \[ z-w=re^{i\theta}, \] write \[ z=e^{i\theta}a,\quad w=e^{i\theta}b \] where \(a,b\) are real numbers.
Then, \[ zw=e^{2i\theta}ab \] But from Step 3, \[ zw=e^{2i\theta} \] Therefore, \[ ab=1 \] Now, \[ z\overline{w} = (e^{i\theta}a)(e^{-i\theta}b) \] \[ z\overline{w}=ab \] Thus, \[ z\overline{w}=1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{z\overline{w}=1} \]