# Obtain $z_1,z_2$ and $z_1/z_2$

Given $$j = \sqrt{-1}$$

$$z_1 = 5 \left(\dfrac{\cos 126^\circ{} + j \sin 126^\circ{}}{\cos 72^\circ{} + j \sin 72^\circ{}}\right);$$

$$z_2 = 2\cos 30^\circ{} + j \sin 30^\circ{};$$

Find using algebraic calculations $$z_1\cdot z_2$$ and $$\frac{z_1}{z_2}$$.

It is advised to use the exponential form to have fewer calculations. The problem is that $$2\cos(30^\circ{})$$. It is not a mistake in fact the $$r_2$$ is given in the results as $$\left(\frac{\sqrt {13}}{2}\right)$$ so the $$2$$ is actually only multiplying $$\cos x.$$

• z1 = 5 *(cos 126° + j sin 126°) / (cos 72° + j sin 72°); r2 = sqrt(13)/2 sorry for not posting the correct way the math formule Jan 1 '20 at 18:00
• Hello, welcome to Math Stack Exchange! Your question seems a bit confusing. is $j$ mean to be $\sqrt{-1}$? Jan 1 '20 at 18:01
• You can, and you should, post your formulas using mathjax, otherwise they are unreadable. Jan 1 '20 at 18:01
• Does $j={}{}i$? Jan 1 '20 at 19:05
• yes $j = i = \sqrt(-1)$ Jan 1 '20 at 21:29

Hint ($$i=\sqrt {-1}$$): $$z_1=5\frac{e^{\frac{7\pi} {10}i}}{e^{\frac{4\pi} {10}i}}=5e^{\frac{3\pi} {10}i};\quad z_2=r_2e^{i\phi_2},$$ where $$r_2=\frac {\sqrt {13}}2;\quad\phi_2=\arctan\frac1 {2\sqrt3}.$$
Explaination: the general transformation of complex numbers to polar representation reads: $$x+iy=\operatorname{sign}(x)\sqrt{x^2+y^2}e^{i\arctan\frac yx}$$
• $r_2$ is given in the results of the exercise, therefore I cannot just assume it as known and use it to reverse the process. What I would like to understand is, given that the usual formula for writing a complex is $r(cosx + i sinx)$, how is it possible to have in this case $(a*cosx + i sinx)$? Jan 3 '20 at 10:57