Question

How do you convert $$3-3i$$ to polar form?

Answer 1

Hint: 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 plot different functions in the x-y plane, similarly, we can also plot the complex numbers in the form of graphs, and such a plane is known as an argand plane. If a complex number is represented by $$x+iy$$ , then we plot the point $$(x,y)$$ in the argand plane. We then join this point with the origin to form the complex vector. The distance from the origin is defined by $$r$$ and the angle of the complex vector with the origin is defined by $$\theta$$ .

Complete step-by-step solution:
Now, starting off with the solution for the given problem, we say that,
For any complex number $$a+ib$$ , $$a$$ represents the real point and $$b$$ 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 $$\tan \theta ={\displaystyle \frac{b}{a}}$$ . The distance between the point from the origin is denoted by $$r$$ and is expressed as, $$r=\sqrt{a^2+b^2}$$ . Now representing the complex number or the point in polar form we get,
$$r{e}^{i\theta }=r(\cos \theta +i\sin \theta )$$ . 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 $$(3,-3)$$ , hence the distance between this point and the origin is expressed as, $$r=\sqrt{a^2+b^2}$$,
$$\begin{array}{rl}& \Rightarrow r=\sqrt{3^2+{(-3)}^2}\\ & \Rightarrow r=\sqrt{9+9}\\ & \Rightarrow r=\sqrt{18}\\ & \Rightarrow r=3\sqrt{2}\end{array}$$
We now calculate the angle, we get
$$\tan \theta ={\displaystyle \frac{b}{a}}$$ , finding the value of $\theta$ ,
$$\theta ={\tan }^{-1}\left({\displaystyle \frac{b}{a}}\right)$$
Putting the respective points, we get,
$$\begin{array}{rl}& \theta ={\tan }^{-1}\left({\displaystyle \frac{-3}{3}}\right)\\ & \Rightarrow \theta ={\tan }^{-1}(-1)\\ & \Rightarrow \theta =-{\displaystyle \frac{\pi }{4}}\end{array}$$
Thus we represent the complex number in polar form as,
$$\begin{array}{rl}& 3\sqrt{2}{e}^{i(-{\displaystyle \frac{\pi }{4}})}\\ & =3\sqrt{2}{e}^{-i{\displaystyle \frac{\pi }{4}}}\end{array}$$

Note: For such problems, we need to remember the general form of complex numbers and how we can plot them on the argand plane. The conversion from general form to polar form and its subsequent steps should also be kept in mind. We need to calculate the distance and the argument to represent a complex number in polar form.