Question

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

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

Hint:

Use \(|z|^2 = z\bar{z}\).
Rearrange the given equation to group terms or factor.
Test the options if direct derivation is difficult. If an option leads to an identity (like \(0=0\)) when substituted into the rearranged equation, it's a valid solution.

Answer 1

The Correct Option is C

Given \(|z|^2w - |w|^2z = z - w\). We know \(|z|^2 = z\bar{z}\) and \(|w|^2 = w\bar{w}\). Substitute these into the equation: \(z\bar{z}w - w\bar{w}z = z - w\) Rearrange the terms: \(z\bar{z}w + w = w\bar{w}z + z\) Factor out \(w\) from LHS and \(z\) from RHS: \(w(z\bar{z} + 1) = z(w\bar{w} + 1)\) \(w(|z|^2 + 1) = z(|w|^2 + 1)\) This doesn't directly lead to the options easily. Let's try another rearrangement from the original equation: \(|z|^2w + w = |w|^2z + z\) \(w( |z|^2 + 1) = z( |w|^2 + 1)\) Consider the given equation again: \(|z|^2w - |w|^2z = z - w\). \(|z|^2w + w = |w|^2z + z\) \(w(1+|z|^2) = z(1+|w|^2)\). This is the same. Let's group terms with z and w: \(|z|^2w + w = |w|^2z + z\) \(w - z = |w|^2z - |z|^2w\) \(w - z = z(w\bar{w}) - w(z\bar{z})\) If \(w-z \neq 0\) (since \(z, w\) are distinct). The original equation was \(|z|^2w - |w|^2z = z - w\). \(-(w-z) = z\bar{z}w - w\bar{w}z\) If we divide by \(zw\) (since \(z,w\) are non-zero): \(\frac{|z|^2}{z} - \frac{|w|^2}{w} = \frac{1}{w} - \frac{1}{z}\) \(\bar{z} - \bar{w} = \frac{1}{w} - \frac{1}{z}\) \(\overline{(z-w)} = \frac{z-w}{zw}\) If \(z-w \neq 0\), we can divide by \(z-w\): \(\frac{\overline{(z-w)}}{z-w} = \frac{1}{zw}\). Let \(X = z-w\). Then \(\frac{\bar{X}}{X} = \frac{1}{zw}\). We know \(X\bar{X} = |X|^2\), so \(\bar{X} = |X|^2/X\). \(\frac{|X|^2/X}{X} = \frac{|X|^2}{X^2} = \frac{1}{zw}\). This is \( \frac{|z-w|^2}{(z-w)^2} = \frac{1}{zw} \). This seems complicated. Let's go back to \(z\bar{z}w - w\bar{w}z = z - w\). \(z\bar{z}w - z = w\bar{w}z - w\) \(z(\bar{z}w - 1) = w(\bar{w}z - 1)\) \(z(\bar{z}w - 1) = w(\overline{z\bar{w}} - 1)\) This form is not useful unless \(z\bar{w}\) is real. Alternative approach from \(|z|^2w - |w|^2z = z - w\): \(|z|^2w - z = |w|^2z - w\) \(z(\frac{|z|^2w}{z} - 1) = w(\frac{|w|^2z}{w} - 1)\) \(z(\bar{z}w - 1) = w(z\bar{w} - 1)\). This is the same as before. If \(z\bar{w} - 1 = 0\), then \(z\bar{w} = 1\). If this is true, then LHS = \(z(1-1)=0\). RHS = \(w(1-1)=0\). So \(0=0\). This means \(z\bar{w}=1\) is a possible solution, provided \(z, w\) are distinct and non-zero. If \(z\bar{w}=1\), then \(\bar{w} = 1/z\). So \(w = 1/\bar{z}\). Let's check if this satisfies the original equation: \(|z|^2(1/\bar{z}) - |1/\bar{z}|^2 z = z - 1/\bar{z}\) \(z\bar{z}(1/\bar{z}) - (1/|z|^2) z = z - 1/\bar{z}\) \(z - z/|z|^2 = z - 1/\bar{z}\) This requires \(z/|z|^2 = 1/\bar{z}\). \(z/(z\bar{z}) = 1/\bar{z} \Rightarrow 1/\bar{z} = 1/\bar{z}\). This is true. So \(z\bar{w}=1\) is a valid solution. This matches option (c). \[ \boxed{z\bar{w} = 1} \]