Complex numbers in trigonometric form: product and quotient

When a complex number $$z$$ is given in polar form and is therefore defined by only $$(|z|,\alpha)$$ it is possible to easily give the so-called trigonometric form. We proceed in the following way:

$$$z=|z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)]$$$

where $$|z|=\sqrt{a^2+b^2} \ $$ and $$ \ \alpha=\arctan\big( \dfrac{b}{a} \big)$$.

Since the Euler formula says that $$ \ e^{i\alpha}=\cos\alpha+i \cdot\sin\alpha \ $$ the trigonometric form can also be written as follows: $$$z=|z|\cdot e^{i\alpha}$$$

This other way of expressing a complex number gives us a new way of expressing the inverse element of a complex, which is:

$$$ z^{-1}=|z|^{-1}\cdot(\cos\alpha-i \sin\alpha)=|z|^{-1}\cdot e^{-i\alpha}$$$

For example: $$ 4(\cos(30^\circ) +i\cdot\sin(30^\circ))$$ is the complex number that has module $$4$$ and $$30^\circ$$ is the argument. It is also possible to write it as: $$4\cdot e^{i 30^\circ}$$.

$$ 23(\cos(245^\circ) +i\cdot\sin(245^\circ))$$ is also a complex, but has module $$23$$ and $$245^\circ$$ is the argument. It is also possible to write it as: $$23\cdot e^{i 245^\circ}$$.

Product and quotient of complex numbers in trigonometric form

If we write two complex numbers in trigonometric form we have:

$$z_1=|z_1|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)]$$

$$z_2=|z_2|\cdot[\cos(\beta)+i \cdot\sin(\beta)]$$

When we do the product, we have:

$$$ \displaystyle \begin{array}{rl} z_1\cdot z_2 =& |z_1|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)] \cdot |z_2|\cdot[\cos(\beta)+i \cdot\sin(\beta)] \\ =& |z_1|\cdot |z_2|\cdot [\cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta) \\ &+ \ i\cdot (\cos(\alpha)\cdot\sin(\beta)+\sin(\alpha)\cdot\cos(\beta)) ] \end{array} $$$

Knowing the formula of the cosine of the sum and the sine of the sum we obtain:

$$$ z_1\cdot z_2 = |z_1|\cdot |z_2|\cdot [\cos(\alpha+\beta) + i\cdot\sin(\alpha+\beta)] $$$

By applying the formula we multiply: $$$ \big( 2\cdot[\cos(36^\circ)+i\cdot\sin(36^\circ)] \big) \cdot \big( 5\cdot[\cos(120^\circ)+i\cdot\sin(120^\circ)] \big)= $$$ $$$ =2\cdot 5\cdot [\cos(36^\circ+120^\circ)+i\cdot\sin(36^\circ+120^\circ)]$$$

Calculating the stated operations we obtain: $$$ \big( 2\cdot[\cos(36^\circ)+i\cdot\sin(36^\circ)] \big) \cdot \big( 5\cdot[\cos(120^\circ)+i\cdot\sin(120^\circ)] \big)= $$$ $$$ =10\cdot [\cos(156^\circ)+i\cdot\sin(156^\circ)]= 10\cdot e^{i156^\circ}$$$

and these are the final two ways of expressing the result in trigonometric form.

Therefore, from the previous definition, we can say that the module of the product of two complex numbers is equal to the product of the modules of the factors and the argument of the product is equal to the sum of the arguments. It is the same result as in polar form.

Let's observe that the product of two conjugate complex numbers is a real number, since the sum of the arguments is zero.

For example:

Since the angle that forms the complementary of a complex that has argument $$\alpha$$ is always $$360^\circ -\alpha$$, we have:

$$$ \big( 2\cdot[\cos(20^\circ)+i\cdot\sin(20^\circ)] \big) \cdot \big( 2\cdot[\cos(340^\circ)+i\cdot\sin(340^\circ)] \big)= $$$ $$$ = 2\cdot 2 \cdot [\cos(360^\circ)+i\cdot\sin(360^\circ)]= 4\cdot (1+i\cdot0)= 4$$$

So the result is a real number, as the imaginary part is $$$\sin(\alpha+\beta)= \sin(\alpha+360^\circ - \alpha)= \sin(360^\circ)=0$$$

always zero because the angles of two conjugate complex numbers is always $$360^\circ$$ (if we add them).

Given two complex numbers there is always a third number, called quotient, which when multiplied by the second one (not equal to zero) gives a product equal to the first number. Using the product's properties, we see that the quotient will be a number which module is the quotient of the modules and which argument will be the difference of the two arguments, as with the polar form. That's why we are going to use the following formula to solve quotients:

$$$ \dfrac{z_1}{z_2}=\dfrac{|z_1|\cdot[\cos(\alpha)+i\cdot \sin(\alpha)]}{|z_2|\cdot[\cos(\beta)+i\cdot \sin(\beta)]}=\dfrac{|z_1|}{|z_2|}\cdot[\cos(\alpha-\beta)+i\cdot \sin(\alpha-\beta)]$$$

This comes from multiplying and dividing by the conjugate of the above mentioned number, as with the polar form.

Let's see an example: $$$ \dfrac{z_1}{z_2}=\dfrac{7\cdot[\cos(330^\circ)+i\cdot \sin(330^\circ)]}{4\cdot[\cos(50^\circ)+i\cdot \sin(50^\circ)]}=\dfrac{7}{4}\cdot[\cos(280^\circ)+i\cdot \sin(280^\circ)]=\dfrac{7}{4}\cdot e^{280^\circ i}$$$