If |z₁+z₂|²=|z₁|²+|z₂|², where z₁ and z₂ are two complex numbers, then

Correct Answer is : z1z2{z₁}{z₂}z2z1 is purely imaginary

{z₁}{z₂} is purely imaginary

Correct Answer is : z1z2{z₁}{z₂}z2z1 is purely imaginary

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AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

|z_{1}+z_{2}|^{2}=|z_{1}|^{2}+|z_{2}|^{2}

z_{1}

z_{2}

are two complex numbers, then

Solution:

|z_{1}+z_{2}|^{2}=|z_{1}|^{2}+|z_{2}|^{2}

given ...(i)

Let

z_{1}=x+i y, z_{2}=a+i b

Then

\frac{z_{1}}{z_{2}}=\frac{x+i y}{a+i b}=\frac{x+i y}{a+i b} \times \frac{a-i b}{a-i b}

\frac{z_{1}}{z_{2}}=\frac{(a x+y b)+i(a y-x b)}{a^{2}+b^{2}}

...(ii)

From Eq. (i), we get

LHS =|z_{1}+z_{2}|^{2}=|x+a+i(y+b)|^{2}

=(x+a)^{2}+(y+b)^{2}

=x^{2}+a^{2}+2 a x+y^{2}+b^{2}+2 b y

RHS =|x+i y|^{2}+|a+i b|^{2}

=x^{2}+y^{2}+a^{2}+b^{2}

\Rightarrow 2(a x+b y)=0 \Rightarrow a x+b y=0

\text{Put } a x+b y=0 \; \text{ in Eq. (ii)}

\frac{z_{1}}{z_{2}}=\frac{i(a y-x b)}{a^{2}+b^{2}},

which is purely imaginary.

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