First slide

Algebra of complex numbers

Question

If $|\frac{z_{1}}{z_{2}}| = 1 and \arg (z_{1} z_{2}) = 0$ , then

Moderate

A

z₁ = z₂
B

${|z_{2}|}^{2} = z_{1} z_{2}$
C

$z_{1} z_{2} = 1$
D

none of these

Solution

Let $z_{1} = |z_{1}| (\cos θ_{1} + isin θ_{1})$ .
Now, $|\frac{z_{1}}{z_{2}}| = 1 \Rightarrow |z_{1}| = |z_{2}|$
Also, $\arg (z_{1} z_{2}) = 0 or \arg (z_{1}) + \arg (z_{2}) = 0$
$\begin{array}{l} \Rightarrow \arg (z_{2}) = - θ_{1} \\ \Rightarrow z_{2} = |z_{2}| (\cos (- θ_{1}) + isin (- θ_{1})) \\ = |z_{1}| (\cos θ_{1} - isin θ_{1}) = {\bar{z}}_{1} \\ \Rightarrow z_{2} = \bar{({\bar{z}}_{1})} = z_{1} \\ \Rightarrow {|z_{2}|}^{2} = z_{1} z_{2} \end{array}$

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