First slide

Algebra of complex numbers

Question

If for complex numbers z₁ and z₂, arg $(z_{1}) - \arg (z_{2}) = 0$ then $|z_{1} - z_{2}|$ is equal to

Moderate

A

$|z_{1}| + |z_{2}|$
B

$|z_{1}| - |z_{2}|$
C

$| | z_{1} | - | z_{2} | |$
D

0

Solution

We have,
${|z_{1} - z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 |z_{1}| |z_{2}| \cos (θ_{1} - θ_{2})$
where $θ_{1} = \arg (z_{1}) and θ_{2} = \arg (z_{2})$ Given,
$\arg (z_{1} - z_{2}) = 0$
$\begin{array}{l} \Rightarrow {|z_{1} - z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 |z_{1}| |z_{2}| \\ = {(|z_{1}| - |z_{2}|)}^{2} \\ \Rightarrow |z_{1} - z_{2}| = | | z_{1} | - | z_{2} | | \end{array}$

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