AP EAMCET · Maths · Complex Number

If ${|z_{1} + z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2}$ , where $z_{1}$ and $z_{2}$ are two complex numbers, then

A $\frac{z_{1}}{z_{2}}$ is purely real
B $\frac{z_{1}}{z_{2}}$ is purely imaginary
C $\arg (\frac{z_{1}}{z_{2}}) = \frac{π}{4}$
D $|\frac{z_{1}}{z_{2}}| = 1$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{z_{1}}{z_{2}}$ is purely imaginary

Step-by-step Solution

Detailed explanation

Given that: z1+z22=z12+z22, where z1 and z2 are two complex numbers z1+z22=z12+z22+2z1z2cos(x-y), where x & y are arguments of z1 & z2. Here, 2z1z2cos(x-y)=0 ⇒cos(x-y)=0 ⇒x-y=(2n-1)π2 where n∈I Now,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If z1+z22=z12+z22, where z1 and z2 are two complex numbers, then

Step-by-step Solution

If ${|z_{1} + z_{2}|}^{2} = {|z_{1}|}^{2} + {|z_{2}|}^{2}$ , where $z_{1}$ and $z_{2}$ are two complex numbers, then