Question

How do you express the complex number in trigonometric form: $- 6i?$

Answer 1

Find the magnitude of the complex number and then find its argument.
Argument of a complex number $(a + ib)$ is given by ${\tan ^{ - 1}}\dfrac{b}{a}$
And magnitude can be calculated as $\sqrt {{a^2} + {b^2}}$
First convert the complex number into its polar form and then use Euler’s equation to convert the polar form of the complex number into trigonometric form.

Completed step by step solution:
To express a complex number $(a + ib)$ in trigonometric form,
i.e.$r(\cos \theta + i\sin \theta )$ , we have to first find the polar form of the given complex number.

Polar form of a complex number $(a + ib)$ is written as
$= r{e^{i\theta }}$ Where “r” is the magnitude of the complex number, i.e. $r = \sqrt {{a^2} + {b^2}}$
and $\theta$ is the argument of the complex number which is equals to ${\tan ^{ - 1}}\dfrac{b}{a}$

In the given complex number $- 6i$ we can see that the value of $a\;{\text{and}}\;b$ are
$0\;{\text{and}}\; - 6$ respectively.

Now finding the value of “r” in order to write the magnitude of the complex number
$\Rightarrow r = \sqrt {{a^2} + {b^2}} \\\ \Rightarrow r = \sqrt {{0^2} + {{( - 6)}^2}} \\\ \Rightarrow r = \sqrt {36} \\\ \Rightarrow r = 6 \\\$

Since magnitude accepts only positive values so we left the negative value of $\sqrt {36}$

Now we will find the argument of the complex number
$\Rightarrow \theta = {\tan ^{ - 1}}\dfrac{b}{a} \\\ \Rightarrow \tan \theta = \dfrac{b}{a} \\\ \Rightarrow \tan \theta = \dfrac{{ - 6}}{0} \\\$

Which is not defined and we know that tangent function is not defined at $2n\pi \pm \dfrac{\pi }{2},\;{\text{where}}\;n \in I$ , but we can see that this complex number lies at negative y-axis. So the
argument will be $2n\pi - \dfrac{\pi }{2},\;{\text{where}}\;n \in I$

Now writing $- 6i$ in polar form as $6{e^{i\left( { - \dfrac{\pi }{2}} \right)}} = 6{e^{ - \dfrac{{i\pi }}{2}}}$

Remember the Euler’s equation which can be written as
${e^{i\theta }} = \cos \theta + i\sin \theta$

So now writing ${e^{ - \dfrac{{i\pi }}{2}}}$ in trigonometric form using Euler’s formula
${e^{ - \dfrac{{i\pi }}{2}}} = \cos \left( {\dfrac{{ - \pi }}{2}} \right) + i\sin \left( {\dfrac{{ - \pi }}{2}} \right)$

$\therefore$ the required trigonometric form of $- 6i = r\left( {\cos \left( {\dfrac{{ - \pi }}{2}} \right) \+ i\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right) = 6\cos \dfrac{{ - \pi }}{2} + 6i\sin \dfrac{{ - \pi }}{2}$

Note: Check the quadrant after finding the argument because in trigonometry in a complete period of $2\pi$ there exist two equal values for any argument of trigonometric functions, so checking the quadrant will solve this problem.