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Mathematics Question on Algebra of Complex Numbers

Let z1z_1 and z2z_2 be complex numbers such that z1≠z2z_1 \neq z_2 and ∣z1∣=∣z2∣| z_1| = | z_2 | . If Re (z1)>0(z_1) > 0 and Im(z2)<0Im (z_2) < 0 ,then z1+z2z1−z2\frac{z_1+ z_2}{z_1 - z_2} is

A

One

B

real and positive

C

real and negative

D

purely imaginary

Answer

purely imaginary

Explanation

Solution

Z1 & Z2 are complex numbers such that;

Z1≠Z2 and |Z1| = |Z2|

Z1 has positive real part &

Z2 has negative imaginary part
Given, ∣z1∣=∣z2∣\left|z_{1}\right|=\left|z_{2}\right|
⇒∣z1∣2=∣z22∣\Rightarrow \left|z_{1}\right|^{2}=\left|z_{2}^{2}\right|
⇒z1z1ˉ=z2z2ˉ\Rightarrow z_1\bar{z_1} = z_2\bar{z_2}
Now, (z1+z2z1−z2)+(z1+z2‾z1−z2)\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\overline{z_1+z_2}}{z_1-z_2}\right)
=(z1+z2z1−z2)+(zˉ1+zˉ2zˉ1−zˉ2)=\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}\right)
=z1zˉ1+z2zˉ1−z1zˉ2−z2zˉ2+z1zˉ1+z1zˉ2−z2zˉ1+z2zˉ2(z1−z2)(zˉ1−zˉ2)=\frac{z_{1} \bar{z}_{1}+z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{2}+z_{1} \bar{z}_{1}+z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_{2} \bar{z}_{2}}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}
=2(∣z1∣2−∣z2∣2)(z1−z2)(zˉ1−zˉ2)=0(∵∣z1∣2=∣z2∣2)=\frac{2\left(\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=0 \left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)
=z1+z2z1−z2=\frac{z_{1}+z_{2}}{z_{1}-z_{2}} is purely imaginary.