Question

How do you convert \[1-\left( \sqrt{3} \right)i\] to polar form?

Answer 1

Hint: Problems like these are basically a simple demonstration of complex numbers and its application. We need to know that complex numbers can be represented and expressed in different forms. The general form of representing a complex number is \[a+ib\], where the first term is the real part and the second term is the imaginary part. Complex numbers can also be represented in the polar form as \[r{{e}^{i\theta }}\] . Representing any function in the coordinate plane is equivalent to representing a complex number in the argand plane. We have the x and y axis in coordinate geometry, whereas we have real and imaginary axes for complex numbers. The representation of \[a+ib\] is the basic form and the form \[r{{e}^{i\theta }}\] is the polar form. We can also 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. If we join this point with the origin, it represents the complex vector in the argand plane. 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 (with respect to the origin) 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 \[\left( 1,-\sqrt{3} \right)\] , hence the distance between this point and the origin is expressed as, \[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\],
\[\begin{align}
  & \Rightarrow r=\sqrt{{{1}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}} \\
 & \Rightarrow r=\sqrt{1+3} \\
 & \Rightarrow r=\sqrt{4} \\
 & \Rightarrow r=2 \\
\end{align}\]
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,
\[\begin{align}
  & \theta ={{\tan }^{-1}}\left( \dfrac{-\sqrt{3}}{1} \right) \\
 & \Rightarrow \theta =-{{\tan }^{-1}}\left( \sqrt{3} \right) \\
 & \Rightarrow \theta =-\dfrac{\pi }{3} \\
\end{align}\]
Thus we represent the complex number in polar form as,
\[2{{e}^{-i\dfrac{\pi }{3}}}\]

Note: In problems like these, the most important thing is to remember the various representational forms of complex numbers. We must keep in mind that the distance of the point in the argand plane from the origin gives us the value of ‘r’ and the angle between the complex vector and the real axis gives us the argument. After calculating the value of ‘r’ and ‘theta’, we need to put these values in \[r{{e}^{i\theta }}\] to get the required polar form.