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Mathematics Question on Complex Numbers and Quadratic Equations

If for complex numbers (z1(z_1 and z2) z_2) arg (z1−z2)(z_1 - z_2) = 0, then ∣z1−z2∣| z_1 - z_2| is equal to

A

∣z1∣+∣z2∣|z_1| + |z_2|

B

∣z1∣−∣z2∣|z_1| - |z_2|

C

∣ ∣z1∣−∣z2∣ ∣|\,|z_1| - |z_2|\,|

D

00

Answer

∣ ∣z1∣−∣z2∣ ∣|\,|z_1| - |z_2|\,|

Explanation

Solution

We have ∣z1−z2∣2=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣cos\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-2\left|z_{1}\right|\left|z_{2}\right| cos (θ1−θ2)\left(\theta_{1}-\theta_{2}\right) where θ1=arg.z1,θ2=arg.z2\theta_{1}=arg. z_{1}, \theta_{2}=arg. z_{2} Since arg.∣z1−z2∣=0arg. \left|z_{1}-z_{2}\right|=0 ∴∣z1−z2∣2=∣z1∣2+∣z2∣2−2∣z1∣∣z2∣\therefore \left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-2\left|z_{1}\right|\left|z_{2}\right| =(∣z1∣−∣z2∣2)=\left(\left|z_{1}\right|-\left|z_{2}\right|^{2}\right) ∴∣z1−z2∣=∣∣z1∣−∣z2∣∣\therefore \left|z_{1}-z_{2}\right|=|\left|z_{1}\right|-\left|z_{2}\right||