# Find $\frac{z_2}{z_1}$ in the form $a+bi$

Let $$z_1=\cos\frac{7\pi}{16}+i\sin\frac{7\pi}{16}$$ and $$z_2=\cos\frac{15\pi}{16}+i\sin\frac{15\pi}{16}$$

For this problem im asked to write $\frac{z_2}{z_1}$ in the form $a+bi$, so essentially to simplify and then go from polar form to Cartesian form. In this problem im not allowed to use a calculator or any other technology to aid me. The problem is that I dont know enough about trigonometric properties or algebraic manipulation to simplify it any further than the initial statment $$\frac{\cos\frac{15\pi}{16}+i\sin\frac{15\pi}{16}}{\cos\frac{7\pi}{16}+i\sin\frac{7\pi}{16}}$$

I know this question is vague and it seems like there's no work put into it(there is), but any hints or ideas for what to do next would be appreciated

• $\cos\phi + i\sin\phi = e^{i\phi}$. Aug 20 '18 at 17:09

$$\frac{\cos\frac{15\pi}{16}+i\sin\frac{15\pi}{16}}{\cos\frac{7\pi}{16}+i\sin\frac{7\pi}{16}}\cdot\frac{\cos\frac{7\pi}{16}-i\sin\frac{7\pi}{16}}{\cos\frac{7\pi}{16}-i\sin\frac{7\pi}{16}}=\frac{\cos(\frac{15\pi}{16})\cos(\frac{7\pi}{16})+\sin(\frac{15\pi}{16})\sin(\frac{7\pi}{16})+i\left (\sin(\frac{15\pi}{16})\cos(\frac{7\pi}{16})-\sin(\frac{7\pi}{16})\cos(\frac{15\pi}{16})\right)}{1}$$

Now use that:

• $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
• $\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)$

Can you take it from here?

• I hadnt thought of multiplying the denominator by its conjugate. inally solved it. Cheers! Aug 20 '18 at 17:54

Hint: Use $e^{i\theta}=\cos\theta+i\sin\theta$, in case you don't know the proof, see here.

Hint: Note that $$\cos(\frac{7\pi}{6})+i\sin(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}-\frac{1}{2}i$$

$$\frac{\cos\frac{15\pi}{6}+i\sin\frac{15\pi}{6}}{\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}}=\dfrac{\exp(\frac{15\pi}{6}i)}{\exp(\frac{7\pi}{6}i)}=\exp(\frac{8\pi}{6}i)=\exp(\pi i)\exp\left(\frac{\pi}{3}i\right)=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$$