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

z1=x+iy z2=a+ib |z1+z2|2=|z1|2+|z2|2⇒(x+a)2+(y+b)2=x2+y2+a2+b2 ⇒ax+by=0 z1z2=ax+bya2+b2+(ay−bxa2+b2)i =ay−bxa2+b2iPurely imaginary

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

z1=x+iy z2=a+ib |z1+z2|2=|z1|2+|z2|2⇒(x+a)2+(y+b)2=x2+y2+a2+b2 ⇒ax+by=0 z1z2=ax+bya2+b2+(ay−bxa2+b2)i =ay−bxa2+b2iPurely imaginary

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${| z_{1} + z_{2} |}^{2} = {| z_{1} |}^{2} + {| z_{2} |}^{2}$ where z₁, z₂ are two complex numbers then

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$\frac{z_{1}}{z_{2}}$ is purely real

$\frac{z_{1}}{z_{2}}$ is purely Imaginary

are $(\frac{z_{1}}{z_{2}}) = \frac{π}{4}$

$|\frac{z_{1}}{z_{2}}| = 1$

answer is B.

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Detailed Solution

$z_{1} = x + i y z_{2} = a + i b {| z_{1} + z_{2} |}^{2} = {| z_{1} |}^{2} + {| z_{2} |}^{2}$
$\Rightarrow {(x + a)}^{2} + {(y + b)}^{2} = x^{2} + y^{2} + a^{2} + b^{2} \Rightarrow a x + b y = 0 \frac{z_{1}}{z_{2}} = \frac{a x + b y}{a^{2} + b^{2}} + (\frac{a y - b x}{a^{2} + b^{2}}) i = (\frac{a y - b x}{a^{2} + b^{2}}) i$
Purely imaginary