If z-0- z2-0-z1-z2- and arg z1- z2- then-

The correct option is C = 0 Let z_1 = re^iθ,|z_1| = | z_2 | and arg z_1 + argz_2 = π Then z_2 = re^(π θ)i ∴z̅_̅2̅ = re^ i(π θ) = e^iθ e^ iπ = e^iθ ∴ z_1 ...

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Byju's Answer

Standard XI

Mathematics

Equations of the form acos?+bsin? = c

If z|=0, z2|=...

Question

If $z_{1} \neq 0, z_{2} \neq 0, | z_{1} |$ = $| z_{2} |$ and arg $z_{1} +$ arg $z_{2}$ = $π$ , then:

= 0

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Solution

The correct option is A
= 0

Let $z_{1}$ = $r e^{i θ}, | z_{1} |$ = $| z_{2} |$ and arg $z_{1}$ + arg $z_{2}$ = $π$

Then $z_{2}$ = $r e^{(π - θ) i}$

$∴_{2}$ = $r e^{- i (π - θ)}$ = $e^{i θ} e^{- i π}$ = $- e^{i θ}$

$∴ z_{1} +_{2}$ = $z_{1} - z_{1}$ = 0

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If z1≠0,z2≠0,|z1| = |z2| and arg z1+ arg z2 = π, then:

The correct option is A = 0 Let z1 = reiθ,|z1| = |z2| and arg z1 + argz2 = π Then z2 = re(π−θ)i ∴¯z2 = re−i(π−θ) = eiθe−iπ = −eiθ ∴z1+¯z2 = z1−z1 = 0

If $z_{1} \neq 0, z_{2} \neq 0, | z_{1} |$ = $| z_{2} |$ and arg $z_{1} +$ arg $z_{2}$ = $π$ , then:

The correct option is A
= 0

Let $z_{1}$ = $r e^{i θ}, | z_{1} |$ = $| z_{2} |$ and arg $z_{1}$ + arg $z_{2}$ = $π$

Then $z_{2}$ = $r e^{(π - θ) i}$

$∴_{2}$ = $r e^{- i (π - θ)}$ = $e^{i θ} e^{- i π}$ = $- e^{i θ}$

$∴ z_{1} +_{2}$ = $z_{1} - z_{1}$ = 0