If - z 1- z 2- and z 1- z 2- then z 1- z 2 is equal toA- 0B- Purely imaginaryC- Purely realD- None of these

We have arg(z_1/z_2) = π ⇒ arg z_1 arg z_2 = π ⇒ arg z_1 = π + arg z_2 Let arg (z_2) = θ, then arg (z_1) = π + θ ∴ z_1 = |z_1| [cos (π +θ)+isin (π +θ)] = ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If | z 1|=| z...

Question

If $| z_{1} |$ = $| z_{2} |$ and arg $(\frac{z_{1}}{z_{2}})$ = $π$ , then $z_{1} + z_{2}$ is equal to

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Purely imaginary

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Purely real

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None of these

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Solution

The correct option is A
0

We have arg $(\frac{z_{1}}{z_{2}})$ = $π$

$\Rightarrow$ $a r g z_{1} - a r g z_{2}$ = $π$

$\Rightarrow$ $a r g z_{1}$ = $π + a r g z_{2}$

Let $a r g (z_{2})$ = $θ$ , then $a r g (z_{1})$ = $π + θ$

$∴$ $z_{1}$ = $| z_{1} | [c o s (π + θ) + i s i n (π + θ)]$

= $| z_{1} | [- c o s θ - i s i n θ]$

And $z_{2}$ = $| z_{2} | . (c o s θ + i s i n θ)$ = $| z_{1} | (c o s θ + i s i n θ)$ = $- z_{1}$

$∴ z_{1} + z_{2}$ = $z_{1} - z_{1}$ = 0

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If |z1| = |z2| and arg(z1z2) = π, then z1+z2 is equal to

The correct option is A 0 We have arg(z1z2) = π ⇒ argz1−argz2 = π ⇒ argz1 = π+argz2 Let arg(z2) = θ, then arg(z1) = π+θ ∴ z1 = |z1|[cos(π+θ)+isin(π+θ)] = |z1|[−cosθ−isinθ] And z2 = |z2|.(cosθ+isinθ) = |z1|(cosθ+isinθ) = −z1 ∴z1+z2 = z1−z1 = 0

If $| z_{1} |$ = $| z_{2} |$ and arg $(\frac{z_{1}}{z_{2}})$ = $π$ , then $z_{1} + z_{2}$ is equal to