Complex Numbers 2 Question 40
41. For complex numbers $z$ and $w$, prove that $|z|^{2} w-|w|^{2} z=z-w$, if and only if $z=w$ or $z \bar{w}=1$.
$(1999,10 M)$
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Solution:
- Given, $|z|^{2} w-|w|^{2} z=z-w$
$\Rightarrow z \bar{z} w-w \bar{w} z=z-w$
Taking modulus of both sides, we get
$$ \begin{aligned} & |z w||\bar{z}-\bar{w}|=|z-w| \\ & \Rightarrow \quad|z w||\bar{z}-\bar{w}|=|\overline{z-w}| \quad[\because|z|=|\bar{z}|] \\ & \Rightarrow \quad|z w||\bar{z}-\bar{w}|=|\bar{z}-\bar{w}| \\ & \Rightarrow \quad|\bar{z}-\bar{w}|(|z w|-1)=0 \\ & \Rightarrow \quad|\overline{z-w}|=0 \quad \text { or } \quad|z w|-1=0 \\ & \Rightarrow \quad|z-w|=0 \quad \text { or } \quad|z w|=1 \\ & \Rightarrow \quad z-w=0 \quad \text { or } \quad|z w|=1 \\ & \Rightarrow \quad z=w \text { or } \quad|z w|=1 \end{aligned} $$
Now, suppose $z \neq w$
Then, $|z w|=1$ or $|z||w|=1$
$\Rightarrow \quad|z|=\frac{1}{|w|}=r$
[say]
Let
$$ z=r e^{i \theta} \text { and } \quad w=\frac{1}{r} e^{i \varphi} $$
On putting these values in Eq. (i), we get
$$ \begin{array}{rlrl} & & r^{2} \frac{1}{r} e^{i \varphi}-\frac{1}{r^{2}}\left(r e^{i \theta}\right) & =r e^{i \theta}-\frac{1}{r} e^{i \varphi} \\ \Rightarrow \quad r e^{i \varphi}-\frac{1}{r} e^{i \theta} & =r e^{i \theta}-\frac{1}{r} e^{i \varphi} \\ \Rightarrow & r+\frac{1}{r} e^{i \varphi} & =r+\frac{1}{r} e^{i \theta} \\ \Rightarrow & & e^{i \varphi} & =e^{i \theta} \Rightarrow \varphi=\theta \end{array} $$
Therefore, $\quad z=r e^{i \theta}$ and $w=\frac{1}{r} e^{i \theta}$
$$ \Rightarrow \quad z \bar{w}=r e^{i \theta} \cdot \frac{1}{r} e^{-i \theta}=1 $$
NOTE ‘If and only if’ means we have to prove the relation in both directions.
Conversely
Assuming that $z=w$ or $z \bar{w}=1$
If
$$ \begin{aligned} z & =w, \text { then } \\ \text { LHS } & =z \bar{z} w-w \bar{w} z=|z|^{2} \cdot z-|w|^{2} \cdot z \\ & =|z|^{2} \cdot z-|z|^{2} \cdot z=0 \end{aligned} $$
and $\quad$ RHS $=z-w=0$
If $\quad z w=1$, then $\bar{z} \bar{w}=1$ and
$$ \text { LHS }=z \bar{z} w-w \bar{w} z=\bar{z} \cdot 1-\bar{w} \cdot 1 $$
$$ =\bar{z}-\bar{w}=z-\bar{w}=0=RHS $$
Hence proved.
Alternate Solution
We have,
$$ |z|^{2} w-|w|^{2} z=z-w $$
$$ \begin{aligned} \Leftrightarrow & & |z|^{2} w-|w|^{2} z-z+w & =0 \\ & \Leftrightarrow & \left(|z|^{2}+1\right) w-\left(|w|^{2}+1\right) z & =0 \\ & \Leftrightarrow & \left(|z|^{2}+1\right) w & =\left(|w|^{2}+1\right) z \\ & \Leftrightarrow & \frac{z}{w} & =\frac{|z|^{2}+1}{|w|^{2}+1} \end{aligned} $$
$\therefore \frac{z}{w}$ is purely real.
$$ \frac{\bar{z}}{\bar{w}}=\frac{z}{w} \quad \Rightarrow \quad z \bar{w}=\bar{z} w $$
$$ \begin{array}{lc} \text { Again, } & |z|^{2} w-|w|^{2} z=z-w \\ \Leftrightarrow & z \cdot \bar{z} w-w \cdot \bar{w} z=z-w \\ \Leftrightarrow & z(\bar{z} w-1)-w(z \bar{w}-1)=0 \\ \Leftrightarrow & (z-w)(z \bar{w}-1)=0 \\ \Leftrightarrow & z=w \text { or } z \bar{w}=1 \end{array} $$
Therefore, $|z|^{2} w-|w|^{2} z=z-w$ if and only if $z=w$ or $z \bar{w}=1$.
