$\displaystyle\left|\left(i^3 – \sqrt3\right)\cdot\left(\cos\frac\pi 7+i\sin\frac\pi 7\right)\right|$
I'm not sure how should I start doing this. Am I supposed to take real and imaginary parts of $\left(i^3 – \sqrt3\right)$ and $\left(\cos\frac\pi 7+i\sin\frac\pi 7\right)$ separately and then find absolute value?
Best Answer
Note that $i^3-\sqrt 3 =-2(\frac{\sqrt 3}{2}+i\frac{1}{2})=2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
$x=i^3-\sqrt 3 =-2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})=2e^{i\frac{\pi}{6}}$ $y=\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}=e^{i\frac{\pi}{7}}$
$|xy|=2|e^{i(\frac{\pi}{6}+ \frac{\pi}{7})}|=2$