Complex Numbers: Definition, Properties, Formulas and Examples

Complex Numbers: If someone asks you to find the square root of \(16,\) you would say it will be \(4\) but if you are asked the square root of \(-16,\) what will be your answer? There is no real number whose square is \(-16,\) but there exists a complex number whose square can be negative. We just define a quantity \(i = \sqrt { – 1} \) such that its square is \( – 1.\) so, \(\sqrt { – 16}  = 4i\).

With this simple concept, we can solve a lot of lengthy mathematical problems in a simpler way. Apart from this, complex numbers also play an important role in signal processing,\(AC\) circuit analysis, Quantum mechanics, etc. It helps in solving difficult problems with ease. Let’s learn about complex numbers in detail.

What is a Complex Number?

Complex Numbers are the numbers that can be written in the form of \(x + iy,\) where \(x,\,y\) are real numbers and \(i = \sqrt { – 1} \) Here,\(i\) is an imaginary number whose square is \( – 1.\) The set of complex numbers is represented by \(C.\)

If \(z\,\, = \,a\, + ib\) is the complex number, then \(a\) and \(b\) are called real and imaginary parts,respectively, of the complex number and written as \({\mathop{\rm Re}\nolimits} \,(z)\, = \,a,\,{\mathop{\rm Im}\nolimits} \,(z)\, = \,b\)

For example:
In \(z\, = \,3\, + \,i4\,,{\mathop{\rm Re}\nolimits} \,(z)\, = \,3,{\mathop{\rm Im}\nolimits} \,(z) = 4\)
and if \(z = \,3 = 3 + 0i\) is also a complex number as it can be written in the form of \(x\, + iy\,\) This means all real numbers come under complex numbers.

The relations “greater than” and “less than” are not defined for complex numbers with the imaginary part. For example, we can’t compare \(2 + \,3i\) and \(1\, + \,2i\)

If the imaginary part of a complex number is \(0\) then the complex number is known as a purely real number, and if the real part is \(0\), then it is called a purely imaginary number, for example, \(1\) is a purely real number because its imaginary part is \(0\) and \(8i\) is a purely imaginary number because its real part is \(0\).

Two complex numbers \({z_1}\, = \,a\,\, + \,\,ib\) and \({z_2}\, = \,c\,\, + \,\,id\) are said to be equal if \(a\,\, = \,\,c\) and \(b\,\, = \,\,d\).

A complex number \(z = \,a\,\, + \,\,ib\,\) can be represented by a unique point \(P\,(a,\,\,b)\)in the cartesian plane referred to as a pair of rectangular axes. The complex number \(0\,\, + \,\,0i\) represent the origin \(O(0,\,0).\)A purely real number \(a\), i.e.,\((a\,\, + \,\,0i)\) is represented by the point \((a,\,\,0)\) on \(x – \) axis is called a real axis. A purely imaginary number \(ib,i.e.,(0\,\, + \,\,ib)\) is represented by the point \((0,\,\,b)\) on \(y – \)axis. Therefore, \(y – \)axis is called an imaginary axis

Similarly, the representation of complex numbers as points in the plane is known as the Argand diagram. The plane representing complex numbers as points is called the complex plane or Argand plane.

It is to be noted that
\({i^{4n}}\, = 1\)
\({i^{4n + 1}}\, = \,i\)
\({i^{4n + 2}} =  – 1\)
\({i^{4n + 3}}\, = \, – i\)
Where,
\(n\) is an integer

Properties of Complex Numbers

Some of the properties of complex numbers which help to solve a lot of problems are as follows:

1. Addition and Subtraction of complex numbers: Let \({z_1}\, = \,a\,\, + \,\,ib\) and \({z_2}\, = \,c\,\, + \,\,id\) be two complex numbers then \({z_1} \pm {z_2}\, = \,(a\, + \,c)\, \pm \,i\,(b\, + \,d)\)

2. Multiplication of complex numbers: Let \({z_1}\, = \,a\,\, + \,\,ib\) and \({z_2}\, = \,c\,\, + \,\,id\) be two complex numbers then \({z_1}.\,{z_2} = (a\,\, + \,\,ib)(c\,\, + \,\,id)\, = \,(ac\, – bd) + \,i\,(ad\, + \,bc)\)

3. Conjugate of a complex number: Let \(z\, = \,a\,\, + \,\,ib\) be a complex number. Then a complex number obtained by changing the sign of imaginary part of the complex number is called the conjugate of \(z\) and it is denoted by \(\overline z \,,\,i.e.,\overline z \, = \,a\, – \,ib\). We have:

(i) \(\overline {(\overline z )}  = z\)
(ii) \(z + \overline z \, = 2{\mathop{\rm Re}\nolimits} \,(z),\,z\, – \overline z  = 2i{\mathop{\rm Im}\nolimits} (z)\)
(iii) \(z\, = \,\overline z \, = \,0\, \Leftrightarrow \,z\)
(iv) \(z\,.\,\overline z \, = \,{\left\{ {{\mathop{\rm Re}\nolimits} (z)} \right\}^2}\, + \,{\left\{ {{\mathop{\rm Im}\nolimits} (z)} \right\}^2}\)
(v) \((\overline {{z_1} + {z_2}} )\, = \,\overline {{z_1}} \, + \,\overline {{z_2}} ,\,(\overline {\overline {{z_1} – {z_2}} )} \, = \overline {{z_1}}  – \overline {{z_2}} \)
(vi) \((\overline {{z_1}\,.\,{z_2}} )\, = \,(\overline {{z_1}} )(\overline {{z_2}} ),\frac{{{z_1}}}{{{z_2}}} = \frac{{(\overline {{z_1}} )}}{{(\overline {{z_2}} )}}; (\overline {{z_2}} \, \ne \,0)\)

4. Modulus of a Complex Number: Let \(z = \,a\,\, + \,\,ib\) be a complex number. Then the modulus(absolute value) of \(z\) is denoted by \(\left| {\overline z } \right| = \sqrt {{a^2} + {b^2}} \)

(i) \(\left| z \right|\, = \,0\, \Leftrightarrow \,z\, = \,0\,i.e.,{\mathop{\rm Re}\nolimits} (z)\, = \,0\) and \({\mathop{\rm Im}\nolimits} (z)\,\, = \,\,0\)
(ii) \(\left| z \right|\, = \,\left| {\overline z } \right| = \left| { – z} \right|\)
(iii) \( – \left| z \right| \le \,{\mathop{\rm Re}\nolimits} \,(z)\, \le \,\left| z \right|\) and \( – \left| z \right|\, \le \,{\mathop{\rm Im}\nolimits} \,(z)\, \le \,\left| z \right|\)
(iv) \(z\overline z = {\left| z \right|^2} = {\left| {\overline z } \right|^2}\)
(v) \(\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|,\,\left| {{z_2}} \right|,\,\left| {\frac{{{z_1}}}{{{z_2}}}} \right|;\,\left( {{z_2} \ne 0} \right)\)
(vi) \({\left| {{z_1}\, + \,{z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + 2{\mathop{\rm Re}\nolimits} ({z_1}{\overline z _2})\)
(vii) \({\left| {{z_1}\, – \,{z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} – 2{\mathop{\rm Re}\nolimits} ({z_1}{\overline z _2})\)
(viii) \(\left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|\)
(ix) \(\left| {{z_1} – {z_2}} \right| \ge \left| {{z_1}} \right| – \left| {{z_2}} \right|\)
(x) \({\left| {a{z_1}\, – \,b{z_2}} \right|^2} + \,{\left| {b{z_1}\, + \,a{z_2}} \right|^2} = ({a^2}\, + \,{b^2})({\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2})\)
In particular \({\left| {{z_1}\, – \,{z_2}} \right|^2} + {\left| {{z_1} + {z_2}} \right|^2} = 2({\left| {{z_1}} \right|^2}\, + \,{\left| {{z_2}} \right|^2})\)

The multiplicative inverse (reciprocal) of a complex number \(z = a + ib\,\left( { \ne 0} \right)\) is \(\frac{1}{z} = \frac{{a – ib}}{{{a^2} + \,{b^2}}}\, = \,\frac{{\overline z }}{{{{\left| z \right|}^2}}}\)

Mathematical Applications of Complex Numbers

Some examples given below are solved with the concept of complex numbers. Complex numbers problems are based on:

a) Comparing the real and imaginary part
Example 1: What will be the value of \(x\) and \(y\) if \((x\, + \,2y)\, + \,i(2x\, – \,y) = 3\) where \(x,y \in \,R\)
Solution: Since, \((x + 2y) + i(2x – y) = 3\)
\( \Rightarrow (x\, + \,2y) + \,i\,(2x\, – \,y) = 3\)
Comparing real part and imaginary part from the both side
\( \Rightarrow \,x + \,2y = 3,2x – y = 0\)
\( \Rightarrow x = 1,\,y = 2\)

b) Involving iota

Example 2: What will be the value of \({i^{34}} + {i^{35}} + {i^{36}} + {i^{37}}?\)
Solution: we know that
\({i^{4n}} = 1\)
\({i^{4n + 1}} = i\)
\({i^{4n + 2}} =  – 1\)
\({i^{4n + 3}} =  – i\)
\(n\) is an integer
Hence,
\({i^{34}} + {i^{35}} + {i^{36}} + {i^{37}}\)
\( \Rightarrow  – 1 – i + 1 + i = 0\)

c) Solving Quadratic Equations
Example 3: Solve \(\sqrt 5 {x^2} + x + \sqrt 5 = 0\)
Solution: Here, the discriminant of the equation is
\({1^2} – 4 \times \sqrt 5  \times \sqrt 5  = 1 – 20 =  – 19\)
Therefore, the solutions are
\(x = \frac{{ – 1 \pm \sqrt { – 19} }}{{2\sqrt 5 }} = \frac{{ – 1 \pm \sqrt {19} i}}{{2\sqrt 5 }}\)

d) Use of Rationalisation

Example 4: If \((1 + i)z = (1 – i)\overline z ,\) then show that \(z\, =  – i\overline z \)
Solution: We have,
\((1\, + \,i)z\, = \,(1 – i)\overline z \)
\( \Rightarrow z = \frac{{1 – i}}{{1 + i}}\overline z  = \frac{{(1 – i)(1 – i)}}{{(1 + i)(1 – i)}}\overline z  = \frac{{{{(1 – i)}^2}}}{{(1 – {i^2})}}\overline z  = \frac{{1 – 2i + {i^2}}}{{1 + 1}}\overline z  = \frac{{1 – 2i + {i^2}}}{{1 + 1}}\overline z  =  – i\overline z \)

e) Putting the value of \(z = \,x\,\, + \,\,iy\)

Example 5: If \(z = \,x\,\, + \,\,iy\) then show that \(z\overline z \, + 2(z\, + \,\overline z ) + b = 0\) where \(b\, \in R\) represents a circle.
Solution.: Given that, \(z\, = \,x\, + \,iy\,\, \Rightarrow \,\overline z  = x – iy\)
Now, \(z\overline z \, + \,2(z\, + \,\overline z )\, + \,b = 0\)
\( \Rightarrow \,(x\, + \,iy)(x – \,iy)\, + 2(x + iy + x – iy) + b = 0\)
\( \Rightarrow \,{x^2} + {y^2} + 4x\, + b = 0;\) this is the equation of a circle

Example 6: If the real part of \(\frac{{\overline z  + 2}}{{\overline z  – 1}}\) is \(4\) then show that the locus of the point representing \(z\) in the complex plane is a circle.
Solution: Let: \(z = \,x + \,iy\)
Now, \(\frac{{\overline z  + 2}}{{\overline z  – 1}} = \frac{{x – iy + 2}}{{x – iy – 1}} = \frac{{\left[ {(x + 2) – iy} \right]\left[ {(x – 1) + iy} \right]}}{{\left[ {(x – 1) – iy} \right]\left[ {(x – 1) + iy} \right]}}\)
\( = \frac{{(x – 1)(x + 2) + {y^2} + i\left[ {(x + 2)} \right]y – (x – 1)y}}{{{{(x – 1)}^2} + {y^2}}}\)
\(\Rightarrow \frac{{(x – 1)(x + 2) + {y^2}}}{{{{(x – 1)}^2} + {y^2}}} = 4 \Rightarrow {x^2} + x – 2 + {y^2} = 4({x^2} – 2x + 1 + {y^2})\)
\( \Rightarrow \,3{x^2} + 3{y^2} – 9x + 6 = 0\) which represents a circle. Hence, locus of \(z\) is a circle

Solved Examples – Complex Numbers

Q.1. Express \({(5 – 3i)^3}\) in the form of \(a\,\, + \,\,ib\)
Ans: \({\left( {5 – 3i} \right)^3} = {\left( 5 \right)^3} – {\left( {3i} \right)^3} – 3{\left( 5 \right)^2}3\left( i \right) + 3\left( 5 \right)3{\left( i \right)^2}\)
\( = 125\, – \,27{i^3} – (25)(9i) + 15(9{i^2})\)
\( = 125 + 27i\, – \,225i – 135\)
\( = \, – 10 – 198i\)
Hence, the value of \({(5 – 3i)^3}\) is \( – 10 – 198i\)

Q.2. If \({(x + iy)^{\frac{1}{3}}} = a + ib\) where \(x,y,a,b \in R,\) show that \(\frac{x}{a} – \frac{y}{b} =  – 2({a^2} + {b^2})\)
Ans: \({(x + iy)^{\frac{1}{3}}} = a + ib\)
\( \Rightarrow \,x + iy\, = {(a + ib)^3}\)
\(i.e.,x + iy = {a^3} + {i^3}{b^3} + 3iab(a + ib)\)
\( = {a^3} – i{b^3} + i3{a^2}b – 3a{b^2}\)
\( = {a^3} – 3a{b^2} + i(3{a^2}b – {b^3})\)
\( \Rightarrow x = {a^3}\, – 3a{b^2}\) and \(y = \,3{a^2}b – {b^3}\)
Thus, \(\frac{x}{a} = {a^2} – 3{b^2}\) and \(frac{y}{b} = 3{a^2} – {b^2}\)
So, \(\frac{x}{a} – \frac{y}{b} = {a^2} – 3{b^2} – 3{a^2} + {b^2} =  – 2{a^2} – 2{b^2} =  – 2({a^2} + {b^2})\)

Q.3. If the imaginary part of  \(\frac{{2z + 1}}{{iz + 1}}\,is\, – 2\) then show that the locus of the point representing \(z\) in the argand plane is a straight line.
Ans: Let \(z = x + iy.\) Then
\(\frac{{2z + 1}}{{iz + 1}}\, = \,\frac{{2(x + iy) + 1}}{{i(x + iy) + 1}} = \frac{{(2x + 1) + i2y}}{{(1 – y) + ix}}\)
\( = \frac{{\left\{ {(2x + 1) + i2y} \right\}}}{{\left[ {(1 – y) + ix} \right]}} \times \frac{{\left[ {(1 – y) – ix} \right]}}{{\left\{ {(1 – y) – ix} \right\}}}\)
\( = \frac{{(2x + 1 – y) + i(2y – 2{y^2} – 2{x^2} – x)}}{{1 + {y^2} – 2y + {x^2}}}\)

Thus,\({\mathop{\rm Im}\nolimits} \left( {\frac{{2z + 1}}{{iz + 1}}} \right) = \frac{{2y – 2{y^2} – 2{x^2} – x}}{{1 + {y^2} – 2y + {x^2}}}\)
but \({\mathop{\rm Im}\nolimits} \left( {\frac{{2z + 1}}{{iz + 1}}} \right) =  – 2\) Given
So, \(\frac{{2y – 2{y^2} – 2{x^2} – x}}{{1 + {y^2} – 2y + {x^2}}} =  – 2\)
\( \Rightarrow 2y – 2{y^2} – 2{x^2} – x =  – 2 – 2{y^2} + 4y – 2{x^2}\)
\(i.e.,x + 2y – 2 = 0\) which is the equation of a line

Q.4. If \(\left| {{z^2} – 1} \right| = {\left| z \right|^2} + 1\) then show that \(z\) lies on an imaginary axis.
Ans: Let \(z = x + iy\)
Then, \(\left| {{z^2} – 1} \right| = {\left| z \right|^2} + 1\)
\( \Rightarrow \left| {{x^2} – {y^2} – 1 + i2xy} \right| = {\left| {x + iy} \right|^2} + 1\)
\( \Rightarrow {({x^2} – {y^2} – 1)^2} + 4{x^2}{y^2} = {({x^2} + {y^2} + 1)^2}\)
\( \Rightarrow 4{x^2} = 0\,i.e.,x = 0\)
Hence \(z\) lies on \(y\)-axis

Q.5. Find the value of \(\left| {\frac{{(1 + i)(2 + i)}}{{3 + i}}} \right|\)
Ans: We have given that,
\(\left| {\frac{{(1 + i)(2 + i)}}{{3 + i}}} \right| = \frac{{(\left| {1 + i} \right|\left| {2 + i} \right|)}}{{\left| {3 + i} \right|}}\)
We can write it as,
\( = \frac{{\sqrt {{{(1)}^2} + {{(1)}^2}} \sqrt {(2) + {{(1)}^2}} }}{{\sqrt {{{(3)}^2} + {{(1)}^2}} }}\)
On solving, we get, \( = \frac{{\sqrt {1 + 1} \sqrt {4 + 1} }}{{\sqrt {9 + 1} }}\,= \frac{{\sqrt 2 \sqrt 5 }}{{\sqrt {10} }}\)
We can write it as \( \frac{{\sqrt 2 \sqrt 5 }}{{\sqrt 2 \sqrt 5 }}\, = 1\)

Summary

In this article, we have learnt about the Complex Numbers, its properties and its examples and got to know that it plays an essential role in the mathematical world. Complex numbers are impossible to ignore in the scientific world, despite the fact that humans cannot physically visualise them.

Complex numbers exemplify the importance of mathematics in science, as it serves as both a powerful language for describing complex phenomena and a full toolkit for solving challenging issues.

You can also refer to the NCERT Solutions for Maths provided by academic experts at Embibe for your final or board exam preparation.

Frequently Asked Questions (FAQ) – Complex Numbers

Q.1. Why are complex numbers important?
Ans: Complex numbers plays an important role in signal processing, AC circuit analysis, Quantum mechanics etc. It helps in solving difficult problems with ease. When it comes to modelling systems with sinusoidal inputs, complex numbers offer certain important mathematical qualities that can make your job easier. A good example is electrical circuits. To summarise, you use complex numbers to avoid having to learn Calculus. Even the most difficult Geometrical problems are easily tackled with the concept of complex numbers

Q.2. How to solve complex numbers?
Ans: To solve the complex numbers problems, we have a basic rule, i.e. to put complex number \(z = x + iy\) then we need to simplify it according to the needs of the question given.

Q.3. Is \(0\) a complex number?
Ans: Yes, \(0\,\, = \,\,0\,\, + \,\,i0\) is a complex number because it can be represented in the form of \(a\,\, + \,\,ib\) where \(a\,\, = \,\,0\) and \(b\,\, = \,\,0\)

Q.4. How to multiply complex numbers?
Ans: A complex number is formed by multiplying two complex numbers. In other words, the multiplication of two complex numbers can be written as \(A\,\, + \,\,iB\) where \(A\) and \(B\) are both real numbers. For example, The multiplication of \({z_1}\, = \,2\, + 3i\) and \({z_2} = 1 + i\) will be \((2 + 3i)(1 + i) = 2 + 2i + 3i – 3 =  – 1 + 5i\)

Q.5. How to find roots of complex numbers?
Ans: We can find the roots of complex numbers easily with the following methods.
1. The first step is to let’s assume that the roots of the complex number are \(a + ib,\) for example \(\sqrt {1 + i}  = a + ib\)
2. Then, we square it both sides and then compare the real part and imaginary part of the equations
3. From the above steps, we would get two equations, and we have two variables as well.
4. Solve these two equations, then the values of \(a\) and \(b\) are obtained, which should be verified from the previous steps.