In the set of complex numbers let $z_{1}=\operatorname{cis}\left(\dfrac{\pi}{7}\right)$ 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
Best Answer
$$z_1+z_2 = \cos(\pi/7)+2 +i (\sin(\pi/7)+1)$$
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 $\sqrt{x^2+y^2}$.