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. $$\dfrac{56_{180^{\circ}}}{23_{77^{\circ}}}=\Big(\dfrac{56}{23}\Big)_{180^{\circ}-77^{\circ}}=\Big(\dfrac{56}{23}\Big)_{103^{\circ}}$$

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

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

Solution:

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

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

Hide solution and development

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: $$160_{300^{\circ}}:(4_{76^{\circ}}\cdot 12_{12^{\circ}})=160_{300^{\circ}}:48_{88^{\circ}}={\dfrac{160}{48}}_{300^{\circ}-88^{\circ}} = {\dfrac{10}{3}}_{212^{\circ}}$$

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

Solution:

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

2) $$$\begin{array}{l} z_1=3_{\frac{130}{3}^{\circ}} \\ z_2= 3_{\frac{490}{3}^{\circ}} \\ z_3=3_{\frac{850}{3}^{\circ}} \end{array} $$$

Hide solution and development