Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Presumably $cis(\theta) = e^{i\theta}=\cos(\theta) +i \sin(\theta)$ – Henry Feb 23 '12 at 23:21
up vote 2 down vote accepted

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


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

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 $\sqrt{x^2+y^2}$.

share|cite|improve this answer
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 $\cos^2(t) + \sin^2(t)=1$.

share|cite|improve this answer

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


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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.