Problems from Operations with complex numbers in polar form

Development:

These operations are a quotient of complex numbers in polar form. Dividim els mòduls and restem els arguments. We divide the modules and reduce the arguments.

1. ${\displaystyle \frac{{56}_{{180}^{\circ }}}{{23}_{{77}^{\circ }}}}=({\displaystyle \frac{56}{23}}{)}_{{180}^{\circ }-{77}^{\circ }}=({\displaystyle \frac{56}{23}}{)}_{{103}^{\circ }}$

   In this case we have left the module indicated with the fraction because the result is not an integer.

2. ${\displaystyle \frac{{24}_{{60}^{\circ }}}{4_{{60}^{\circ }}}}={{\displaystyle \frac{24}{4}}}_{{60}^{\circ }-{60}^{\circ }}=6_{0^{\circ }}$.

Solution:

1. ${\displaystyle \frac{{56}_{{180}^{\circ }}}{{23}_{{77}^{\circ }}}}=({\displaystyle \frac{56}{23}}{)}_{{103}^{\circ }}$

2. ${\displaystyle \frac{{24}_{{60}^{\circ }}}{4_{{60}^{\circ }}}}=6_{0^{\circ }}$.

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Development:

1) The first we will do is the product: $4_{{76}^{\circ }}\cdot {12}_{{12}^{\circ }}=(4\cdot 12{)}_{{76}^{\circ }+{12}^{\circ }}={48}_{{88}^{\circ }}$

And now we can do the division: ${160}_{{300}^{\circ }}:(4_{{76}^{\circ }}\cdot {12}_{{12}^{\circ }})={160}_{{300}^{\circ }}:{48}_{{88}^{\circ }}={{\displaystyle \frac{160}{48}}}_{{300}^{\circ }-{88}^{\circ }}={{\displaystyle \frac{10}{3}}}_{{212}^{\circ }}$

2) $\sqrt[3]{{27}_{{130}^{\circ }}}={\sqrt[3]{27}}_{\frac{{130}^{\circ }+{360}^{\circ }k}{3}}=3_{\frac{{130}^{\circ }+{360}^{\circ }k}{3}}$ for $k=0,1,2$.

This way, $${\displaystyle \sqrt[3]{{27}_{{130}^{\circ }}}=\{\begin{array}{l}{3}_{{\frac{130}{3}}^{\circ }}\\ 3_{{\frac{490}{3}}^{\circ }}\\ 3_{{\frac{850}{3}}^{\circ }}\end{array}}$$

Solution:

1) ${{\displaystyle \frac{10}{3}}}_{{212}^{\circ }}$

2) $$\begin{array}{l}{z}_1=3_{{\frac{130}{3}}^{\circ }}\\ z_2=3_{{\frac{490}{3}}^{\circ }}\\ z_3=3_{{\frac{850}{3}}^{\circ }}\end{array}$$

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