If $\left|Z1\right|=\left|Z2\right|\text{}\text{and}\text{}argz1\sim argz2=\pi$ , how do you show that $Z1+Z2=0$ ?

elvishwitchxyp
2022-01-17
Answered

If $\left|Z1\right|=\left|Z2\right|\text{}\text{and}\text{}argz1\sim argz2=\pi$ , how do you show that $Z1+Z2=0$ ?

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temzej9

Answered 2022-01-18
Author has **30** answers

The proof uses Euler’s Identity, actually its reciprocal: ${e}^{-i\pi}=-1$ .

Let$r=\left|{z}_{1}\right|=\left|{z}_{2}\right|\text{}\text{and}\text{}\theta =arg\left({z}_{1}\right)\text{}\text{so}\text{}arg\left({z}_{2}\right)=arg\left({z}_{1}\right)-\pi =\theta -\pi$ .

We get$z}_{1}=r{e}^{i\theta$

$z}_{2}=r{e}^{i(\theta -\pi )}=r{e}^{i\theta}{e}^{-i\pi}=-r{e}^{i\theta$

${z}_{1}+{z}_{2}=r{e}^{i\theta}\pm r{e}^{i\theta}=0$

Let

We get

godsrvnt0706

Answered 2022-01-19
Author has **31** answers

Let $\left|z1\right|=\left|z2\right|=r$

$arg\left(z1\right)=\theta$

$Arg\left(z2\right)=\varphi$

$z}_{1}=r{e}^{i\theta$ ……(1)

$z2=r{e}^{i\varphi}$ ………(2)

And$\theta -\varphi =\pi$

${e}^{ix}=\mathrm{cos}x+i\mathrm{sin}x$

${e}^{i(\pi +x)}=\mathrm{cos}(\pi +x)+i\mathrm{sin}(\pi +x)$

$e}^{i(\pi +x)}=-\mathrm{cos}x-i\mathrm{sin}x=-(\mathrm{cos}x+i\mathrm{sin}x)=-{e}^{ix$

Here$\theta =\pi +\varphi$

So this reduces to

$e}^{i\theta}=-{e}^{i\varphi$

Multiplying by r on both sides

$r{e}^{i\theta}=-r{e}^{i\varphi}$

From equations (1) and (2)

The above expression becomes

$z1=-z2$

That is

$z1+z2=0$ .

And

Here

So this reduces to

Multiplying by r on both sides

From equations (1) and (2)

The above expression becomes

That is

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