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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 $\cos x.$

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  • $\begingroup$ 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 $\endgroup$
    – Phi Party
    Jan 1 '20 at 18:00
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    $\begingroup$ Hello, welcome to Math Stack Exchange! Your question seems a bit confusing. is $j$ mean to be $\sqrt{-1}$? $\endgroup$ Jan 1 '20 at 18:01
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    $\begingroup$ You can, and you should, post your formulas using mathjax, otherwise they are unreadable. $\endgroup$
    – Lee Mosher
    Jan 1 '20 at 18:01
  • $\begingroup$ Does $j={}{}i$? $\endgroup$ Jan 1 '20 at 19:05
  • $\begingroup$ yes $j = i = \sqrt(-1)$ $\endgroup$
    – Phi Party
    Jan 1 '20 at 21:29
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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: $$ x+iy=\operatorname{sign}(x)\sqrt{x^2+y^2}e^{i\arctan\frac yx} $$

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  • $\begingroup$ $r_2$ 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 $(a*cosx + i sinx)$? $\endgroup$
    – Phi Party
    Jan 3 '20 at 10:57
  • $\begingroup$ I have added an explanation. $\endgroup$
    – user
    Jan 3 '20 at 15:06

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