Trigonometric or Polar Form of a Complex Number: Definition, Representation, and Examples

Trigonometric or Polar Form of a Complex Number: A complex number can be represented in three ways such as geometrical, rectangular, and polar forms. Plotting the complex number \(z = x + iy\) is similar to plotting a real number, except that the horizontal axis represents the real part of the number, \(x\) and the vertical axis represents the imaginary part of the number, \(y\). Note that no matter what form a complex number is written in, they are always interchangeable, using certain parameters.

In this article, let us discuss more the representation of a complex number in polar form, their operations, and the conversion of rectangular form to polar form and vice versa.

Geometrical Form of a Complex Number

A complex number \(z = x + iy\) is represented by a point \((x,\,y)\) on the Argand plane. To represent the complex number \(z = x + iy\) geometrically, we take mutually perpendicular lines \({X^\prime }OX\) and \({Y^\prime }OY\). Now plot a point whose \(x\) and \(y\) coordinates are respectively real and imaginary parts of \(z\). The point \(P\left( {x,\,y} \right)\) represents the complex number \(z = x + iy\).

If a complex number is purely real, then the imaginary part is zero. So, a purely real number is represented by a point on \(x\)-axis. A purely imaginary complex number is represented by a point on \(y\)-axis. That is why \(x\)-axis is known as the real axis and \(y\)-axis is known as the imaginary axis. The plane in which we represent a complex number geometrically is known as the complex plane or Argand plane or Gaussian plane. The point \(P\), plotted on the Argand plane, is called the Argand diagram.

The length of line segment \(OP\) is called the modulus of \(z\) and it is denoted by \(\left| z \right|\).

Therefore, from the figure, we have:

\(O{P^2} = O{M^2} + M{P^2}\) [since \(OMP\) is a right-angled triangle]

\( \Rightarrow O{P^2} = {x^2} + {y^2}\)

\( \Rightarrow OP = \sqrt {{x^2} + {y^2}} \)

Hence, \(|z| = \sqrt {{x^2} + {y^2}}  = \sqrt {{{\left\{ {{\mathop{\rm Re}\nolimits} (z)} \right\}}^2} + {{\left\{ {{\mathop{\rm Im}\nolimits} (z)} \right\}}^2}} \)

The angle \(\theta \) made by \(OP\) with the positive direction of \(x\)-axis in anticlockwise direction is known as the argument or the Amplitude of \(z\) and it is denoted by arg \((z)\) or amp \((z)\)

Therefore, from figure, we have: \(\tan \,\theta = \frac{{PM}}{{OM}}\)

\( \Rightarrow \tan \,\theta = \frac{y}{x}\)

\( \Rightarrow \tan \,\theta = \frac{{{\mathop{\rm Im}\nolimits} (z)}}{{{\mathop{\rm Re}\nolimits} (z)}}\)

\( \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{{{\mathop{\rm Im}\nolimits} (z)}}{{{\mathop{\rm Re}\nolimits} (z)}}} \right)\)

The angle \(\theta \) ha infinitely many values differing by multiples of \(2\pi \). The unique value of \(\theta \) such that, \( – \pi < \theta \le \pi \) is called the principal value of the Amplitude of principal argument. This has a severe drawback because \({z_1} = 1 + i\sqrt 3 \) and \({z_2} = – 1 – i\sqrt 3 \) are two different complex numbers represented by two different points on the Argand plane but their arguments seem to be \({\tan ^{ – 1}}\sqrt 3 = \frac{\pi }{3}\) or \(\frac{{4\pi }}{3}\), which is not correct. In fact, the arguments is the common solution of simultaneous trigonometric equations \(\cos \,\theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\) and \(\sin \,\theta = \frac{y}{{\sqrt {{x^2} + {y^2}} }}\)

Since the above system of equations has infinitely many solutions. Therefore, there can be many arguments of \(z = x + iy\). The argument \(\theta \) which satisfies the inequality \( – \pi < \theta \le \pi \) is usually known as the principal argument of \(z\). The argument \(z\) depends upon the quadrant in which the point \(P\) lies.

Argument or Amplitude of a Complex Number

The polar form of the complex number is represented in the form of modulus and argument. But, before we learn the third form, let’s understand the argument of a complex number.

Case (i): When \(x > 0\) and \(y > 0\)

Since, \(x\) and \(y\) both positive, hence the point \(P(x,\,y)\) representing \(z = x + iy\) in the Argand plane lies in the first quadrant. Let \(\theta \) be the argument of \(z\) and \(\alpha \) be the acute angle satisfying \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\)

Therefore, if \(x\) and \(y\) both are positive, then the argument of \(z = x + iy\) is the acute angle given by \(\tan \,\alpha = \frac{y}{x}\)

Case (ii): When \(x < 0\) and \(y > 0\)

In this case, the point \(P\left( {x,\,y} \right)\) representing \(z = x + iy\) in the Argand plane lies in the second quadrant. Let \(\theta \) be the argument of \(z\) and \(\alpha \) be the acute angle satisfying \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\)

It is evident from the figure that, \(\theta = \pi – \alpha \)

Thus, if \(x < 0\) and \(y > 0\), then the argument of \(z = x + iy\) is \(\pi – \alpha \), where \(\alpha \) is the acute angle given by \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\).

Case (iii): When \(x < 0\) and \(y < 0\)

In this case, the point \(P(x,\,y)\) representing \(z = x + iy\) in the Argand plane lies in the third quadrant. Let \(\theta \) be the argument of \(z\) and \(\alpha \) be the acute angle satisfying \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\).

It is evident from the figure that, \(\theta = – (\pi – \alpha ) = \alpha – \pi \)

Thus, if \(x < 0\) and \(y < 0\), then the argument of \(z = x + iy\) is \(\alpha – \pi \), where \(\alpha \) is the acute angle given by \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\).

Case (iv): When \(x > 0\) and \(y < 0\)

In this case, the point \(P(x,\,y)\) representing \(z = x + iy\) in the Argand plane lies in the fourth quadrant. Let \(\theta \) be the argument of \(z\) and \(\alpha \) be the acute angle satisfying \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\).

It is evident from the figure that, \(\theta = – \alpha \)

Thus, if \(x > 0\) and \(y < 0\), then the argument of \(z = x + iy\) is \( – \alpha \), where \(\alpha \) is the acute angle given by \(\tan \,\alpha = \left| {\frac{y}{x}} \right|\).

Polar Form or Trigonometrical of a Complex Number

The polar form of a complex number is another way of representing the complex number. So usually we represent the complex number in the form \(z = x + iy\), where \(i\) is an imaginary number and \(x,\,y\) are two real numbers. But in polar form, the complex numbers are represented by using modulus and argument. The modulus of a complex number is called absolute value, this polar form is represented with the help of polar coordinates of real and imaginary numbers in the coordinate system.

So, by the geometrical representation of \(z = x + iy\), we get \(OP = |z|\) and \(\angle POX = \theta = \arg (z)\)

From \(\Delta POM\), we have,

\(\cos \,\theta = \frac{{OM}}{{OP}}\)

\( \Rightarrow \cos \,\theta = \frac{x}{{|z|}}\)

\( \Rightarrow x = |z|\cos \,\theta \)

Similarly, \(\sin \,\theta = \frac{{PM}}{{OP}}\)

\( \Rightarrow \sin \,\theta = \frac{y}{{|z|}}\)

\( \Rightarrow y = |z|\sin \,\theta \)

\(\therefore \,z = x + iy\)

\( \Rightarrow z = |z|\cos \,\theta + i|z|\sin \,\theta \)

\( \Rightarrow z = |z|(\cos \,\theta + i\,\sin \,\theta )\)

\( \Rightarrow z = r(\cos \,\theta + i\,\sin \,\theta )\)

where

\(r = \left| z \right|\)

\(\theta = \arg (z)\)

If we use the general value of argument \(\theta \), then the polar form is given by \(z = r[\cos (2n\pi + \theta ) + i\,\sin (2n\pi + \theta )]\), where \(r = |z|,\,\theta = \arg (z)\) and \(n\) is an integer.

Polar Form in Different Quadrants

Let \(|z| = r\) and \(\alpha \) be the acute angle given by \(\tan \,\alpha = \left| {\frac{{{\mathop{\rm Im}\nolimits} (z)}}{{{\mathop{\rm Re}\nolimits} (z)}}} \right|\). Let \(\theta \) be the argument of \(z\).

Converting a Complex Number from Polar Form to Rectangular Form

If a complex number \(z\) has a magnitude \(r\) and argument \(\theta \) then \(z = r(\cos \,\theta + i\,\sin \,\theta )\) is a polar form of a complex number.

So, if we want to convert from polar form to rectangular form \(\left( {z = x + iy} \right)\), consider \(x = r\,\cos \,\theta \) and \(y = r\,\sin \,\theta \), where \(r = \sqrt {{x^2} + {y^2}} \) and \(\theta = {\tan ^{ – 1}}\left( {\frac{y}{x}} \right)\)

Operations on Polar Form of a Complex Number

Addition of Two Complex Numbers

Let \(z = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\) and \(w = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\) be the two complex numbers in polar form, then \(z + w = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right) + {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\)

\( \Rightarrow z + w = \left( {{r_1}\,\cos \,{\theta _1} + {r_2}\,\cos \,{\theta _2}} \right) + i\left( {{r_1}\,\sin \,{\theta _1} + {r_2}\,\sin \,{\theta _2}} \right)\)

Subtraction of Two Complex Numbers

Let \(z = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\) and \(w = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\) be the two complex numbers in polar form, then \(z – w = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right) – {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\)

\( \Rightarrow z – w = \left( {{r_1}\,\cos \,{\theta _1} – {r_2}\,\cos \,{\theta _2}} \right) + i\left( {{r_1}\,\sin \,{\theta _1} – {r_2}\,\sin \,{\theta _2}} \right)\)

Multiplication of Two Complex Numbers

Consider two complex numbers in polar form:

\(z = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\)

\(w = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\)

Multiplying two complex numbers, we get,

\(zw = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right) \times {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\)

\( \Rightarrow zw = {r_1}{r_2}\left[ {\left( {\cos \,{\theta _1}\,\cos \,{\theta _2} – \sin \,{\theta _1}\,\sin \,{\theta _2}} \right) + i\left( {\sin \,{\theta _1}\,\cos \,{\theta _2} + \cos \,{\theta _1}\,\sin \,{\theta _2}} \right)} \right]\)

\( \Rightarrow zw = {r_1}{r_2}\left[ {\cos \left( {{\theta _1} + {\theta _2}} \right) + i\,\sin \left( {{\theta _1} + {\theta _2}} \right)} \right]\)

Quotient of Two Complex Numbers

Consider two complex numbers in polar form:

\(z = {r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)\)

\(w = {r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)\)

Then quotient of these complex numbers is \(\frac{z}{w} = \frac{{{r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)}}{{{r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)}} = \frac{{{r_1}\left( {\cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right)}}{{{r_2}\left( {\cos \,{\theta _2} + i\,\sin \,{\theta _2}} \right)}} \times \frac{{\left( {\cos \,{\theta _2} – i\,\sin \,{\theta _2}} \right)}}{{\left( {\cos \,{\theta _2} – i\,\sin \,{\theta _2}} \right)}}\)

\( = \frac{{{r_1}\left( {\cos \,{\theta _1}\cos \,{\theta _2} – i\,\cos \,{\theta _1}\,\sin \,{\theta _2} + i\,\sin \,{\theta _1}\,\cos \,{\theta _2} – {i^2}\,\sin \,{\theta _1}\,\sin \,{\theta _2}} \right)}}{{{r_2}\left( {{{\left( {\cos \,{\theta _2}} \right)}^2} – {i^2}{{\left( {\sin \,{\theta _2}} \right)}^2}} \right)}}\)

\( = \frac{{{r_1}\left( {\left( {\cos \,{\theta _1}\,\cos \,{\theta _2} + \sin \,{\theta _1}\,\sin \,{\theta _2}} \right) + i\left( {\cos \,{\theta _1}\,\sin \,{\theta _2} – \sin \,{\theta _1}\,\cos \,{\theta _2}} \right)} \right)}}{{{r_2}\left( {{{\left( {\cos \,{\theta _2}} \right)}^2} + {{\left( {\sin \,{\theta _2}} \right)}^2}} \right)}}\)

\( = \frac{{{r_1}}}{{{r_2}}}\left[ {\cos \left( {{\theta _1} – {\theta _2}} \right) + i\,\sin \left( {{\theta _1} – {\theta _2}} \right)} \right],\,{r_2} \ne 0\)

Solved Examples

Q.1. Convert the following complex number \(1 + i\) into polar form.
Ans: Let \(z = 1 + i\)
\( \Rightarrow r = |z| = \sqrt {{1^2} + {1^2}} \)
\(r = \sqrt 2 \)
Let \(\alpha \) be the acute angle given by the formula, then we have \(\tan \,\alpha = \left| {\frac{{{\mathop{\rm Im}\nolimits} (z)}}{{{\mathop{\rm Re}\nolimits} (z)}}} \right|\)
\( \Rightarrow \tan \,\alpha = \frac{1}{1} = 1\)
\( \Rightarrow \tan \,\alpha = \tan \frac{\pi }{4}\)
\( \Rightarrow \alpha = \frac{\pi }{4}\)
The point \(\left( {1,\,1} \right)\) representing the complex number lies in the first quadrant. Therefore, the argument of \(z\) is \(\theta = \alpha = \frac{\pi }{4}\)
Polar form \(z = r(\cos \,\theta + i\,\sin \,\theta )\)
\(z = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\,\sin \frac{\pi }{4}} \right)\)

Q.2. Convert the following complex number \(\frac{{2 + 6\sqrt 3 i}}{{5 + \sqrt 3 i}}\) into polar form.
Ans: Let \(z = \frac{{2 + 6\sqrt 3 i}}{{5 + \sqrt 3 i}}\) and \(r(\cos \,\theta + i\,\sin \,\theta )\) be the polar form of \(z\). Then \(r = |z|\) and \(\theta = \arg (z)\)
\(z = \frac{{2 + 6\sqrt 3 i}}{{5 + \sqrt 3 i}}\)
\( = \frac{{2 + 6\sqrt 3 i}}{{5 + \sqrt 3 i}} \cdot \frac{{(5 – \sqrt 3 i)}}{{(5 – \sqrt 3 i)}}\)
\( = \frac{{28 + 28\sqrt 3 i}}{{28}}\)
\( = 1 + i\sqrt 3 \)
\(\therefore \,r = |z| = \sqrt {1 + 3} = 2\)
Let \(\alpha \) be the acute angle given by \(\tan \,\alpha = \frac{{|{\mathop{\rm Im}\nolimits} (z)|}}{{|{\mathop{\rm Re}\nolimits} (z)|}}\). Then \(\tan \,\alpha = \frac{{\sqrt 3 }}{1} = \sqrt 3 \)
\( \Rightarrow \alpha = \frac{\pi }{3}\)
The point \((1,\,\sqrt 3 )\) representing \(z\) lies in first quadrant. Therefore, \(\theta = \arg (z) = \alpha = \frac{\pi }{3}\).
Hence, the polar form is \(2\left( {\cos \frac{\pi }{3} + i\,\sin \frac{\pi }{3}} \right)\).

Q.3. Convert the following complex number \(\frac{{1 + 7i}}{{{{(2 – i)}^2}}}\) into polar form.
Ans: Let \(z = \frac{{1 + 7i}}{{{{(2 – 1)}^2}}}\)
Let \(r(\cos \,\theta + i\,\sin \,\theta )\) be the polar form of \(z\). Then \(r = |z|\) and \(\theta = \arg (z)\)
\(z = \frac{{1 + 7i}}{{4 – 4i + {i^2}}} = \frac{{1 + 7i}}{{3 – 4i}}\)
\( = \frac{{1 + 7i}}{{3 – 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}\)
\( = \frac{{ – 25 + 25i}}{{25}}\)
\( = – 1 + i\)
\(\therefore \,r = |z| = \sqrt {{{( – 1)}^2} + {{(1)}^2}} = \sqrt 2 \)
Let \(\alpha \) be the acute angle given by \(\tan \,\alpha = \left| {\frac{{{\mathop{\rm Im}\nolimits} (z)}}{{{\mathop{\rm Re}\nolimits} (z)}}} \right|\), then \(\tan \,\alpha = \left| { – \frac{1}{1}} \right| = 1\)
\( \Rightarrow \alpha = \frac{\pi }{4}\)
The point \({\rm{( – 1,}}\,{\rm{1)}}\) representing \(z\) lies in the second quadrant.
Therefore, \(\theta = \arg (z) = \pi – \alpha = \pi – \frac{\pi }{4} = \frac{{3\pi }}{4}\)
Hence, the required polar form is \(z = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\,\sin \frac{{3\pi }}{4}} \right)\).

Q.4. Convert the following complex number \(\frac{{1 + i}}{{1 – i}}\) into polar form.
Ans: Let \(z = \frac{{1 + i}}{{1 – i}}\) 
Let \(r(\cos \,\theta + i\,\sin \,\theta )\) be the polar form of \(z\). Then \(r = |z|\) and \(\theta = \arg (z)\)
Now, \(z = \frac{{1 + i}}{{1 – i}}\)
\( = \frac{{(1 + i)(1 + i)}}{{(1 – i)(1 + i)}}\)
\( = \frac{{1 + 2i + {i^2}}}{{1 – {i^2}}}\)
\( = \frac{{1 + 2i – 1}}{{1 + 1}}\)
\( = i\)
\(\therefore \,z = 0 + 1i\)
\(\therefore \,r = |z| = \sqrt {0 + 1} = 1\)
The point \({\rm{(0,}}\,{\rm{1)}}\) lies in the positive direction of the imaginary axis
\(\therefore \,\arg (z) = \frac{\pi }{2}\)
Therefore, the polar form is
\(z = 1\left( {\cos \frac{\pi }{2} + i\,\sin \frac{\pi }{2}} \right)\)
\(z = \cos \frac{\pi }{2} + i\,\sin \frac{\pi }{2}\)

Q.5. Convert the following polar form \(z = 12\left( {\cos \left( {\frac{\pi }{6}} \right) + i\,\sin \left( {\frac{\pi }{6}} \right)} \right)\) into the rectangular form.
Ans: Given that, the polar form is \(z = 12\left( {\cos \left( {\frac{\pi }{6}} \right) + i\,\sin \left( {\frac{\pi }{6}} \right)} \right)\)
Here, we have \(\cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}\) and \(\sin \left( {\frac{\pi }{6}} \right) = \frac{1}{2}\)
Therefore, \(z = 12\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)\)
\( = (12)\frac{{\sqrt 3 }}{2} + (12)\frac{1}{2}i\)
\( = 6\sqrt 3 + 6i\)
Hence, the required rectangular form is \(6\sqrt 3 + 6i\).

Summary

A complex can be represented in three different forms: geometrical, vectorial, polar, or trigonometrical. The form \(z = a + bi\) is a rectangular form of a complex number, where \(\left( {a,\,b} \right)\) are the rectangular coordinates. The polar form of a complex number is another way of representing a complex number. The horizontal axis is a real axis, and the vertical axis is an imaginary axis. So the polar form of a complex number \(z = a + ib\) is \(z = r(\cos \,\theta + i\,\sin \,\theta )\), where,\(r = |z| = \sqrt {{a^2} + {b^2}} ,\,a = r\,\cos \,\theta \) and \(b = r\,\sin \,\theta \). The various operations performed on the polar form of complex numbers are addition, subtraction, multiplication and quotient. The polar form of a complex number is convertible to other forms and vice versa. Later explained a few of the solved examples based on polar form.

Frequently Asked Questions (FAQs)

Q.2. What is the polar form of a complex number?
Ans: The polar form of a complex number is another way to represent a complex number apart from a rectangular form. The polar form of the complex number \(z = a + ib\) is \(z = r\,\cos \,\theta + ir\,\sin \,\theta \), where \(r\) is the modulus of \(z\), and \(\theta \) is the argument of \(z\).

Q.3. How do you convert to polar form?
Ans: The polar form of a complex number \(z = x + iy\) with coordinates \((x,\,y)\) is given as \(z = r\,\cos \,\theta + ir\,\sin \,\theta = r(\cos \,\theta + i\,\sin \,\theta )\), where \(r = \sqrt {{{(x)}^2} + {{(y)}^2}} \) and \(\theta = {\tan ^{ – 1}}\left( {\frac{y}{x}} \right)\).

Q.4. What is the argument of a complex number?
Ans: The angle formed between the positive \(x\)-axis and the line joining a point with coordinates \((x,\,y)\) of the complex number to the origin is called the argument of the complex number.

Q.5. How do you express complex numbers?
Ans: Complex numbers can be represented in the following ways:
(i) Rectangular form
(ii) Polar form or trigonometric form
(iii) Geometric form

Hope this detailed article on the Trigonometric or Polar Form of a Complex Number helps you in your preparation. In case of any queries, reach out to us in the comment section and we will get back to you at the earliest.