# Sum of two complex numbers

In the set of complex numbers let $z_{1}=\operatorname{cis}(\frac{\pi}{7})$ and $z_{2}=2+i$. Prove that $$|z_{1}+z_{2}|^2=6+4\cos\left(\frac{\pi}{7}\right)+2\sin\left(\frac{\pi}{7}\right)\;.$$

I thought to convert $z_{1}$ into algebric form, because I know how to sum two complex numbers in algebric form. But the argument is not a special angle, easy to find the trig value. Other issue, are the "brackets": I don't know what they mean. Absolute value? Modulus?

Can you explain to me how to do this? Thanks

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Presumably $cis(\theta) = e^{i\theta}=\cos(\theta) +i \sin(\theta)$ –  Henry Feb 23 '12 at 23:21

$$z_1+z_2 = \cos(\pi/7)+2 +i (\sin(\pi/7)+1)$$

so

$$|z_1+z_2|^2 = (\cos(\pi/7)+2)^2 + (\sin(\pi/7)+1)^2$$

so multiply out and use $\cos^2(\theta)+\sin^2(\theta)=1$ to get the desired result.

$|x+iy|$ is the modulus and for real $x$ and $y$ is $\sqrt{x^2+y^2}$.

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You’re missing a square on the absolute value. –  Brian M. Scott Feb 23 '12 at 23:31
@Brian: Thanks - now corrected –  Henry Feb 23 '12 at 23:33

$|\cdot|$ means absolute value: $|a+bi| = \sqrt{a^2 + b^2}$ if $a$ and $b$ are real. Hint: there is nothing special about $\pi/7$. Expand the left side out, and use everybody's favourite trig identity $\cos^2(t) + \sin^2(t)=1$.

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If $z=a+bi$, $|z|=\sqrt{a^2+b^2}$; if $z=re^{i\theta}$, $|z|=|r|$. Pictorially, $|z|$ is just the distance from $z$ to the origin. In your case

$$z_1+z_2=\left(\cos\left(\frac{\pi}7\right)+2\right)+\left(\sin\left(\frac{\pi}7\right)+1\right)i\;,$$

so $$|z_1+z_2|^2=\left(\cos\left(\frac{\pi}7\right)+2\right)^2+\left(\sin\left(\frac{\pi}7\right)+1\right)^2\;,$$ which easily simplifies to the desired result.

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On the last line you have $z^2$ where you want $z_2$ - Muphry's law ;) –  Henry Feb 23 '12 at 23:34
@Henry: Yep. Never fails. –  Brian M. Scott Feb 23 '12 at 23:38
Thank you all:Robert Israel, Henry and Brian M. Scott.Now I understood.Despite the trig or algebric form, we always sum the real with the real part and the imaginary with the imaginary part. –  João Feb 23 '12 at 23:45
@João That is why complex numbers are sometimes represented by vectors starting from the origin: their sums and differences follow the rules for vector addition. –  RecklessReckoner Apr 30 '13 at 2:11