Question

How do you convert $8i$ to polar form?

Answer 1

These types of problems are very straightforward and simple demonstration of complex numbers. We first need to remember and understand how complex numbers can be represented into different forms. The very general form of complex numbers is $a+ib$. Complex numbers can also be represented in the form of $r{{e}^{i\theta }}$ . Just like we can represent different forms of coordinate geometry figures in the x-y plane, we can also represent complex numbers via graphs, and this is plotted in the argand plane. We have the x and y axis in cases of coordinate geometry, and we have real axis and imaginary axis for complex numbers. The representation of $a+ib$ is the general form and the form $r{{e}^{i\theta }}$ is the polar representation form. We can further write ${{e}^{i\theta }}$ as $\cos \theta +i\sin \theta$ .

Complete step-by-step solution:
Now, starting off with the solution for the given problem, we say that,
For any complex number $x+iy$, $x$ represents the real point and $y$ represents the imaginary point. Joining this point with the origin, represents the complex vector. The angle between this vector and the real axis is denoted by $\theta$ , where $\tan \theta =\dfrac{y}{x}$ . The distance between the point from the origin is denoted by $r$ and is expressed as, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ . Now representing the complex number or the point in polar form we get,
$r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$ . Here $r$ is defined as the distance of the complex point from the origin.
Now comparing this equation with our given problem, we can say the complex point is represented as $\left( 0,8 \right)$ , hence the distance between this point and the origin is expressed as, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$,

& \Rightarrow r=\sqrt{{{0}^{2}}+{{8}^{2}}} \\\ & \Rightarrow r=\sqrt{{{8}^{2}}} \\\ & \Rightarrow r=8 \\\ \end{aligned}$$ We now calculate the angle, we get $$\tan \theta =\dfrac{y}{x}$$, finding the value of theta, $$\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$$ Putting the respective points, we get, $$\begin{aligned} & \theta ={{\tan }^{-1}}\left( \dfrac{8}{0} \right) \\\ & \Rightarrow \theta ={{\tan }^{-1}}\left( \infty \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{2} \\\ \end{aligned}$$ Thus we represent the complex number in polar form as, $$8{{e}^{i\dfrac{\pi }{2}}}$$ **Note:** For these kinds of problems, the most important thing to remember is that the different forms we can represent of the complex numbers. We need to keep in mind the distance of the complex point from the origin and the angle the complex vector makes with the real axis in the argand plane. After calculating these things we need to plug in the values into the equation $$r{{e}^{i\theta }}$$ to get the required polar form.