# Operations with complex numbers in polar form

Let's see how to get the product of two complexes that are given in polar form. When we want to multiply two complex numbers occuring in polar form, the modules multiply and the arguments add, giving place to a new complex number. In a general way:

$$\displaystyle \begin{array}{l} z_{1}=|z_{1}|_{\alpha_{1}} \\ z_{2}=|z_{2}|_{\alpha_{2}} \end{array} \left\} \Rightarrow z_{1}\cdot z_{2}=\big( |z_{1}|\cdot |z_{2}| \big)_{\alpha_{1}+\alpha_{2}} \right.$$

For example, $$\displaystyle \begin{array}{l} z_{1}=3_{42^{\circ}} \\ z_{2}=6_{112^{\circ}} \end{array} \left\} \Rightarrow z_{1}\cdot z_{2}=( 3\cdot 6)_{42^{\circ}+112^{\circ}}=18_{154^{\circ}} \right.$$

$$\displaystyle \begin{array}{l} z_{1}=5_{24^{\circ}} \\ z_{2}=4_{90^{\circ}} \end{array} \left\} \Rightarrow z_{1}\cdot z_{2}=( 5\cdot 4)_{24^{\circ}+90^{\circ}}=20_{114^{\circ}} \right.$$

As we see, this way of expressing the complex numbers facilitates and improves the operation of multiplying two complex numbers.

Let's now see what happens with the quotient. If we want to divide two complex numbers in polar form, the procedure to follow is: on the one hand, the modules are divided and, on other one, the arguments are reduced giving place to a new complex number which module is the quotient of modules and which argument is the difference of arguments. In general, it is written as:

$$\displaystyle \begin{array}{l} z_{1}=|z_{1}|_{\alpha_{1}} \\ z_{2}=|z_{2}|_{\alpha_{2}} \end{array} \left\} \Rightarrow \frac{z_{1}}{z_{2}}=\Big( \frac{|z_{1}|}{|z_{2}|} \Big)_{\alpha_{1}-\alpha_{2}} \right.$$

For example, $$\displaystyle \begin{array}{l} z_{1}=14_{89^{\circ}} \\ z_{2}=7_{51^{\circ}} \end{array} \left\} \Rightarrow \frac{z_{1}}{z_{2}}=\Big( \frac{14}{7} \Big)_{89^{\circ}-51^{\circ}} =2_{38^{\circ}} \right.$$

$$\displaystyle \begin{array}{l} z_{1}=35_{354^{\circ}} \\ z_{2}=7_{17^{\circ}} \end{array} \left\} \Rightarrow \frac{z_{1}}{z_{2}}=\Big( \frac{35}{7} \Big)_{354^{\circ}-17^{\circ}} =5_{337^{\circ}} \right.$$

Bearing in mind the product of complex numbers expressed in polar form that we have just seen, we are going to deduce how to work with potencies of complex numbers in polar form.

We already know that a potency to $n$ of a complex number means to multiply the number by itself $n$ times. Since we know that the product in polar form only supposes the product of the modules and the sum of the arguments, the following is deduced:

To find the potency of a complex number in polar form one simply has to do potency asked by the module. The argument, in turn, is affected so that it adds himself the same number of times as the potency we are raising.

Then a new complex number is obtained. It is also in polar form. This in general is written for any complex number as:

$$z=|z|_{\alpha} \ \Rightarrow \ (z)^n = \big(|z|_{\alpha}\big)^n= \big(|z|\stackrel{n}{\cdots} |z|\big)_{\alpha+\stackrel{n}{\dots}+\alpha}= {\big(|z|\big)^{n}}_{n \cdot \alpha}$$

For example, $$z=5_{75^{\circ}} \ \Rightarrow \ (z)^3 = (5_{75^{\circ}})^3= {(5)^3}_{3\cdot75^{\circ}}=125_{225^{\circ}}$$

For example, $$z=2_{30^{\circ}} \ \Rightarrow \ (z)^5= (2_{30^{\circ}})^5= {(2)^5}_{5\cdot30^{\circ}}=32_{150^{\circ}}$$

The $n$-th root of a complex $|R|_{\beta}$ is a complex number $|r|_{\alpha}$ that satisfies:

$$\displaystyle \sqrt[n]{|R|_{\beta}}=|r|_{\alpha} \Leftrightarrow \left\{ \begin{array}{l} |r|=\sqrt[n]{|R|} \\ \alpha=\frac{\beta+2\pi k}{n} \end{array} \right.$$

This is because, concerning the modules, it is apparent that the $n$-th root of one must be the other, and regarding the arguments, $n$ times the sum of one must be the other (as we have already said previously, this is not univocally determined, and that is why we can add up the factor $2\pi$ as many times as we like, and also $360$ degrees which is the same as $2\pi$).

We can write as:

$$z=|z|_{\alpha} \ \Rightarrow \ \sqrt[n]{z}=\sqrt[n]{|z|_{\alpha}}= \big(\sqrt[n]{|z|}\big)_{\frac{\alpha+360^{\circ}k}{n}}$$

For example, $$z=64_{120^{\circ}} \ \Rightarrow \ \sqrt[2]{z}=\sqrt[2]{64_{120^{\circ}}}= (\sqrt[2]{64})_{\frac{120^{\circ}}{2}}=8_{60^{\circ}+\frac{360^{\circ}k}{2}}$$

$$z=36_{250^{\circ}} \ \Rightarrow \ \sqrt[2]{z}=\sqrt[2]{36_{250^{\circ}}}= (\sqrt[2]{36})_{\frac{250^{\circ}}{2}}=6_{125^{\circ}+\frac{360^{\circ}k}{2}}$$

Giving integer values to $k$ from $0$ up to $n-1$ we obtain $n$ arguments that satisfy the condition that we have imposed. For $k$ grater or iqual to $n$ we obtain arguments that differ from the previous ones in an integer of $360^{\circ}$ and consequently they coincide with some of the previous $n$.

Then, let's call $n$-th roots of a complex number $|R|_{\beta}$ to the $n$ complex numbers that take the $n$-th root of the module as their module and as their argument: $\dfrac{\beta+360^{\circ}k}{2}$.

These are, as we have said: $$\sqrt[n]{|R|_{\beta}}= \big(\sqrt[n]{|R|}\big)_{\frac{\beta+360^{\circ}k}{n}}$$

with $k = 0,1,2, \dots, n-1$.

Then for example,

$$\displaystyle \sqrt[3]{8i}=\sqrt[3]{8_{90^{\circ}}}= \left\{ \begin{array}{l} 2_{\frac{90^{\circ}}{3}}=2_{30^{\circ}} \\ 2_{\frac{90^{\circ}+360^{\circ}}{3}}=2_{150^{\circ}} \\ 2_{\frac{90^{\circ}+360^{\circ}\cdot2}{3}}=2_{270^{\circ}} \end{array} \right.$$

These three are all the cubic roots of the complex number, when $k = 0,1,2$ respectively.