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Question: For complex numbers z and w, prove that \({{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w\),...

For complex numbers z and w, prove that ∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w, if and only if z=w or zw‾=1z=w\ or\ z\overline{w}=1.

Explanation

Solution

Hint: We will be using the concept of complex numbers to solve the problem. We will be using the concept that modulus of a complex number ∣z∣2=zz‾{{\left| z \right|}^{2}}=z\overline{z}. Also if a complex number is real then z=z‾z=\overline{z}.

Complete step-by-step answer:
Now, we have been given that we have to prove that ∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w if and only if z=w or zw‾=1z=w\ or\ z\overline{w}=1.
Now, we will take the equation ∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w and prove this is true if and only if z=w or zw‾=1z=w\ or\ z\overline{w}=1.
So, we have,
∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w
On rearranging terms, we have,
∣z∣2w+w=∣w∣2z+z w(1+∣z∣2)=z(1+∣w∣2) wz=1+∣w∣21+∣z∣2 \begin{aligned} & {{\left| z \right|}^{2}}w+w={{\left| w \right|}^{2}}z+z \\\ & w\left( 1+{{\left| z \right|}^{2}} \right)=z\left( 1+{{\left| w \right|}^{2}} \right) \\\ & \dfrac{w}{z}=\dfrac{1+{{\left| w \right|}^{2}}}{1+{{\left| z \right|}^{2}}} \\\ \end{aligned}
Now, since 1+∣w∣21+∣z∣2\dfrac{1+{{\left| w \right|}^{2}}}{1+{{\left| z \right|}^{2}}} is purely real. Therefore, we have,
(wz)=(w‾z)=w‾z wz=w‾z wz‾=zw‾.......(1) \begin{aligned} & \left( \dfrac{w}{z} \right)=\left( \dfrac{\overline{w}}{z} \right)=\dfrac{\overline{w}}{z} \\\ & \dfrac{w}{z}=\dfrac{\overline{w}}{z} \\\ & w\overline{z}=z\overline{w}.......\left( 1 \right) \\\ \end{aligned}
Now, we again use the equation,
∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w
Now, we know that,
∣z∣2=zz‾ ∣w∣2=ww‾ \begin{aligned} & {{\left| z \right|}^{2}}=z\overline{z} \\\ & {{\left| w \right|}^{2}}=w\overline{w} \\\ \end{aligned}
So, we will use this in the equation and so we have,
zz‾w−ww‾z=z−w z‾zw−ww‾z−z+w=0 z‾zw−z+w−ww‾z=0 \begin{aligned} & z\overline{z}w-w\overline{w}z=z-w \\\ & \overline{z}zw-w\overline{w}z-z+w=0 \\\ & \overline{z}zw-z+w-w\overline{w}z=0 \\\ \end{aligned}
Now, we take z and w as common. So, we have,
z(z‾w−1)−w(zw‾−1)=0z\left( \overline{z}w-1 \right)-w\left( z\overline{w}-1 \right)=0
Now, from (1) we have,
z‾w=zw‾\overline{z}w=z\overline{w}
So, using this we have,
z(zw‾−1)−w(zw‾−1)=0 (z−w)(zw‾−1)=0 z=w or zw‾=1 \begin{aligned} & z\left( z\overline{w}-1 \right)-w\left( z\overline{w}-1 \right)=0 \\\ & \left( z-w \right)\left( z\overline{w}-1 \right)=0 \\\ & z=w\ or\ z\overline{w}=1 \\\ \end{aligned}
Hence, we have that,
∣z∣2w−∣w∣2z=z−w{{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w if and only if z=w or zw‾=1z=w\ or\ z\overline{w}=1.

Note: To solve these types of questions it is important to remember that if a complex number z is real that z=z‾z=\overline{z}. Also few other identities to be remember like,
∣z∣2=zz‾{{\left| z \right|}^{2}}=z\overline{z}