If |Z1|=|Z2|\ \text{and}\ arg z1 \sim arg z2=\pi, how do

If , how do you show that $Z1+Z2=0$?
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temzej9
The proof uses Euler’s Identity, actually its reciprocal: ${e}^{-i\pi }=-1$.
Let .
We get ${z}_{1}=r{e}^{i\theta }$
${z}_{2}=r{e}^{i\left(\theta -\pi \right)}=r{e}^{i\theta }{e}^{-i\pi }=-r{e}^{i\theta }$
${z}_{1}+{z}_{2}=r{e}^{i\theta }±r{e}^{i\theta }=0$
Not exactly what you’re looking for?
godsrvnt0706
Let $|z1|=|z2|=r$
$arg\left(z1\right)=\theta$
$Arg\left(z2\right)=\varphi$
${z}_{1}=r{e}^{i\theta }$……(1)
$z2=r{e}^{i\varphi }$………(2)
And $\theta -\varphi =\pi$
${e}^{ix}=\mathrm{cos}x+i\mathrm{sin}x$
${e}^{i\left(\pi +x\right)}=\mathrm{cos}\left(\pi +x\right)+i\mathrm{sin}\left(\pi +x\right)$
${e}^{i\left(\pi +x\right)}=-\mathrm{cos}x-i\mathrm{sin}x=-\left(\mathrm{cos}x+i\mathrm{sin}x\right)=-{e}^{ix}$
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
$z1+z2=0$.