SATHEE: Complex Numbers 2 Question 40

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

$\begin{aligned} | z w | | \bar{z} - \bar{w} | = | z - w | \\ \Rightarrow | z w | | \bar{z} - \bar{w} | = | \overset{―}{z - w} | [∵ | z | = | \bar{z} |] \\ \Rightarrow | z w | | \bar{z} - \bar{w} | = | \bar{z} - \bar{w} | \\ \Rightarrow | \bar{z} - \bar{w} | (| z w | - 1) = 0 \\ \Rightarrow | \overset{―}{z - w} | = 0 or | z w | - 1 = 0 \\ \Rightarrow | z - w | = 0 or | z w | = 1 \\ \Rightarrow z - w = 0 or | z w | = 1 \\ \Rightarrow z = w or | z w | = 1 \end{aligned}$

Now, suppose $z \neq w$

Then, $| z w | = 1$ or $| z | | w | = 1$

$\Rightarrow | z | = \frac{1}{| w |} = r$

[say]

Let

$z = r e^{i θ} and w = \frac{1}{r} e^{i φ}$

On putting these values in Eq. (i), we get

$\begin{aligned} r^{2} \frac{1}{r} e^{i φ} - \frac{1}{r^{2}} (r e^{i θ}) & = r e^{i θ} - \frac{1}{r} e^{i φ} \\ \Rightarrow r e^{i φ} - \frac{1}{r} e^{i θ} & = r e^{i θ} - \frac{1}{r} e^{i φ} \\ \Rightarrow & r + \frac{1}{r} e^{i φ} & = r + \frac{1}{r} e^{i θ} \\ \Rightarrow & e^{i φ} & = e^{i θ} \Rightarrow φ = θ \end{aligned}$

Therefore, $z = r e^{i θ}$ and $w = \frac{1}{r} e^{i θ}$

$\Rightarrow z \bar{w} = r e^{i θ} \cdot \frac{1}{r} e^{- i θ} = 1$

NOTE ‘If and only if’ means we have to prove the relation in both directions.

Conversely

Assuming that $z = w$ or $z \bar{w} = 1$

$\begin{aligned} z & = w, then \\ LHS & = z \bar{z} w - w \bar{w} z = | z |^{2} \cdot z - | w |^{2} \cdot z \\ = | z |^{2} \cdot z - | z |^{2} \cdot z = 0 \end{aligned}$

and RHS $= z - w = 0$

If $z w = 1$ , then $\bar{z} \bar{w} = 1$ and

$LHS = z \bar{z} w - w \bar{w} z = \bar{z} \cdot 1 - \bar{w} \cdot 1$

$= \bar{z} - \bar{w} = z - \bar{w} = 0 = R H S$

Hence proved.

Alternate Solution

We have,

$| z |^{2} w - | w |^{2} z = z - w$

$\begin{aligned} \Leftrightarrow & | z |^{2} w - | w |^{2} z - z + w & = 0 \\ \Leftrightarrow & (| z |^{2} + 1) w - (| w |^{2} + 1) z & = 0 \\ \Leftrightarrow & (| z |^{2} + 1) w & = (| w |^{2} + 1) z \\ \Leftrightarrow & \frac{z}{w} & = \frac{| z |^{2} + 1}{| w |^{2} + 1} \end{aligned}$

$∴ \frac{z}{w}$ is purely real.

$\frac{\bar{z}}{\bar{w}} = \frac{z}{w} \Rightarrow z \bar{w} = \bar{z} w$

$\begin{array}{lc} Again, & | z |^{2} w - | w |^{2} z = z - w \\ \Leftrightarrow & z \cdot \bar{z} w - w \cdot \bar{w} z = z - w \\ \Leftrightarrow & z (\bar{z} w - 1) - w (z \bar{w} - 1) = 0 \\ \Leftrightarrow & (z - w) (z \bar{w} - 1) = 0 \\ \Leftrightarrow & z = w or z \bar{w} = 1 \end{array}$

Therefore, $| z |^{2} w - | w |^{2} z = z - w$ if and only if $z = w$ or $z \bar{w} = 1$ .

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 $z \bar{w} = 1$ .

Solution:

Conversely

Alternate Solution

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Complex Numbers 2 Question 40

41. For complex numbers z and w, prove that |z|2w−|w|2z=z−w, if and only if z=w or zw¯=1.

Solution:

Conversely

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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41. For complex numbers $z$ and $w$ , prove that $| z |^{2} w - | w |^{2} z = z - w$ , if and only if $z = w$ or $z \bar{w} = 1$ .