If |Z1|=|Z2| and argz1∼argz2=π, how do you show that Z1+Z2=0?

Best solutions to - "If |Z1|=|Z2| and argz1∼argz2=π, how do you show"

elvishwitchxyp

2022-01-17

| Z 1 | = | Z 2 | and a r g z 1 \sim a r g z 2 = π

, how do you show that

Z 1 + Z 2 = 0

temzej9

Beginner2022-01-18Added 30 answers

The proof uses Euler’s Identity, actually its reciprocal:

e^{- i π} = - 1

.
Let

r = | z_{1} | = | z_{2} | and θ = a r g (z_{1}) so a r g (z_{2}) = a r g (z_{1}) - π = θ - π

.
We get

z_{1} = r e^{i θ}

z_{2} = r e^{i (θ - π)} = r e^{i θ} e^{- i π} = - r e^{i θ}

z_{1} + z_{2} = r e^{i θ} \pm r e^{i θ} = 0

godsrvnt0706

Beginner2022-01-19Added 31 answers

Let

| z 1 | = | z 2 | = r

a r g (z 1) = θ

A r g (z 2) = ϕ

z_{1} = r e^{i θ}

……(1)

z 2 = r e^{i ϕ}

………(2)
And

θ - ϕ = π

e^{i x} = \cos x + i \sin x

e^{i (π + x)} = \cos (π + x) + i \sin (π + x)

e^{i (π + x)} = - \cos x - i \sin x = - (\cos x + i \sin x) = - e^{i x}

Here

θ = π + ϕ

So this reduces to

e^{i θ} = - e^{i ϕ}

Multiplying by r on both sides

r e^{i θ} = - r e^{i ϕ}

From equations (1) and (2)
The above expression becomes

z 1 = - z 2

That is

z 1 + z 2 = 0

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