If for complex numbers z1 and z2, arg z1−arg⁡z2=0 thenz1−z2 is equal to

We have,z1−z22=z12+z22−2z1z2cos⁡θ1−θ2where θ1=arg⁡z1 and θ2=arg⁡z2 Given,arg⁡z1−z2=0⇒ z1−z22=z12+z22−2z1z2=z1−z22⇒ z1−z2=||z1|−|z2||

If for complex numbers z1 and z2, arg z1−arg⁡z2=0 thenz1−z2 is equal to

We have,z1−z22=z12+z22−2z1z2cos⁡θ1−θ2where θ1=arg⁡z1 and θ2=arg⁡z2 Given,arg⁡z1−z2=0⇒ z1−z22=z12+z22−2z1z2=z1−z22⇒ z1−z2=||z1|−|z2||

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

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

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

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

answer is C.

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Detailed 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,
$\arg (z_{1} - z_{2}) = 0$
$\begin{array}{l} \Rightarrow {|z_{1} - z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 |z_{1}| |z_{2}| \\ = {(|z_{1}| - |z_{2}|)}^{2} \\ \Rightarrow |z_{1} - z_{2}| = | | z_{1} | - | z_{2} | | \end{array}$