# Complex Number and Quadratic Equations - Solutions and Study Material

Complex Number : Let us denote $\sqrt{-1}$ by the symbol i. Then ,we have $i^{2}$ = -1. This means that I is a solution of the equation $x^{2}+1=0$

A number of the form a + ib, where a and b are real numbers, is defined to be a complex .

Complex Number: 2 + 3i, $(-1)+i\sqrt{3}$ , $4+i\left(\frac{-1}{11}\right)$ are complex numbers.

For the Complex Number z = a + ib, a is called the real part, denoted by Re z and b is called the imaginary part denoted by Im z of the complex number z. For example, if z = 2 + i5, then Re z = 2 and Im z = 5.

Two complex numbers $Z_{1}$ = a + ib and $Z_{2}$ = c + id are equal if a = c and b = d

Example: If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

Solution: We have , 4x + i (3x – y) = 3 + i (–6) ... (1)

Equating the real and the imaginary parts of (1), we get ,4x = 3, 3x – y = – 6,

which, on solving simultaneously, give x = $\frac{3}{4}$ , y = $\frac{33}{4}$

Algebra of Complex Numbers: In this Section, we shall develop the algebra of complex numbers

Addition of two complex numbers Let $Z_{1}$ = a + ib and $Z_{2}$ = c + id be any two complex numbers .Then , the $Z_{1}$ is define as follows :

$Z_{1} +Z_{2}$ = (a + c) + i(b + d) , which is again a complex number.

For Example (2 + 3i) +(-6 +5i) = (2-6) + I (3 + 5) = (-4 + 8 i)

The addition of complex number satisfy the following properties:

• The closure law The sum of two complex numbers is a complex number, i.e., $Z_{1} +Z_{2}$ is a complex number for all complex numbers Let $Z_{1}$ and Let $Z_{2}$
• The commutative law For any two complex numbers $Z_{1}$ and $Z_{2}$ ,
$Z_{1} +Z_{2} \:=\:Z_{2} +Z_{1}$
• The associative law For any three complex numbers $Z_{1},Z_{2},Z_{3}$,
$Z_{1} +(Z_{2}+ Z_{3})\:=\: Z_{1}+( Z_{2}+ Z_{3})$
• The existence of additive identity There exists the complex number 0 + i 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number z, z + 0 = z
• The existence of additive inverse To every complex number z = a + ib, we have the complex number – a + i(– b) (denoted as – z), called the additive inverse or negative of z. We observe that z + (–z) = 0 (the additive identity)

## Difference of two Complex Numbers

Given any two complex numbers $Z_{1}\: and\:Z_{2}, Z_{1}- Z_{2}\:= \:Z_{1}+( Z_{2})$

Example: (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i
Solution: (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i

Multiplication of two Complex Numbers: Let $Z_{1}$ = a + ib and $Z_{2}$ = c +id be any any two complex numbers. Then the product $Z_{1} Z_{2}$ is defined as follows :

Example: , (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28

The multiplication of complex numbers possesses the following properties, which we state without proofs.

• The Closure Law: The product of two complex numbers is a complex number the product $Z_{1} Z_{2}$ is a complex number for all complex numbers $Z_{1}\:and\: Z_{2}$
• The Commutative Law: For any two complex numbers $Z_{1}\:and\: Z_{2}, Z_{1} Z_{2}= Z_{2} Z_{1}$
• The Associative Law: For any three complex numbers $Z_{1}, Z_{2}, Z_{3}, \left (Z _{1} Z_{2}\right )Z_{3}\:=\: \left (Z _{2} Z_{3}\right )Z_{1}$
• The Existence of Multiplicative Identity: There exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, for every complex number z.
• The Existence of Multiplicative Inverse: For every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number
• $\frac{a}{a^{2}+b^{2}}+i\frac{-b}{a^{2}+b^{2}}$ (denoted by $\frac{1}{Z}\: or\: Z^{-1}$), called the multiplicative inverse of Z such that . $z\frac{1}{z}$ = 1 (the multiplicative identity)
• The Distributive Law: For any three complex number$z_{1}, z_{2}, z_{3},$
$z_{1}\left (z _{2} +z_{3}\right )\: =\: z_{1}z_{2}+z_{1}z_{3}$
$z_{3}\left (z _{1} +z_{2}\right )\: =\: z_{1}z_{3}+z_{2}z_{3}$

Division of two complex numbers : Given any two complex numbers $z_{1}\:and\:z_{1}$ where , $z_{2}$≠ 0 , the quotient $\frac{z_{1}}{z_{2}}$ is defined by $\frac{z_{1}}{z_{2}}\:=\:z_{1}\frac{1}{z_{2}}$

For Example, let $z_{1}$= 6 + 3 I and $z_{2}$ = 2 – I Then

$\frac{z_{1}}{z_{2}} \:=\:\left ( 6+3i\times \frac{1}{2-i} \right )=6+3i\left ( \frac{2}{2^{2}-(-1^{2})} \right )$

= $\left ( 6+3i \right )\left ( \frac{2+i}{5} \right )=\frac{1}{5}\left [ 12-3+i(6+6) \right ] \:= \:\frac{1}{5}\left ( 9+12i \right )$

Power of i: we know that $i^{3}\: =\: i^{2}i\: =\:-i$

$i^{4}\: =\: (i^{2})^{2}\: =\: -1^{2}\: =\: 1$

$i^{5}\: =\: (i^{2})^{2}i\: =\: -1^{2}i\: =\: i$

The Square Roots of a Negative Real Number: Note that $i^{2}=-1\: and\: (i^{2})\: =\: -1$
Therefore, the square roots of – 1 are i, – i. However, by the symbol $\sqrt{-1}$ we would
mean i only.
Now, we can see that i and –i both are the solutions of the equation $x^{2}+1\: =\: 0\:or\:x^{2}\: =\: -1$
The Modulus and the Conjugate of a Complex Number : Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined to be the non-negative real number $\sqrt{a^{2}+b^{2}}\: i.e\: ,|z|=\sqrt{a^{2}+b^{2}}$ and the conjugate of z, denoted as z, is the complex number a – ib, i.e., z = a – ib

Example: $|3+i|\:=\:\sqrt{3^{2}+1^{2}}\:=\:\sqrt{10}=|2-5i|\:=\:\sqrt{2^{2}+(-5^{2})}\:=\: \sqrt{29}$

$\bar{3+i}\: =\: 3-i,\bar{2-5i}\: =\: 2+i,(\bar{-3i-5})\: =\: 3i-5$

Observe that the multiplicative inverse of the non-zero complex number z is given by

$z^{-1}\:=\:\frac{1}{a+ib}\:=\:frac{a}{a^{2}+b^{2}}+i\frac{-b}{a^{2}+b^{2}}\:=\:\frac{a-ib}{a^{2}+b^{2}}\:= \:\frac{\bar{z}}{|z|^{2}}\:=\:z\bar{z}=|z|^{2}$

Furthermore, the following results can easily be derived. For any two compex numbers z1 and z2 , we have,

• $|z_{1}z_{2}|=|z_{1}||z_{2}|$
• $\frac{|z_{1}|}{|z_{2}|}=|\frac{z_{1}}{z_{2}}|$ provided 2z ≠ 0
• $\bar{z_{1}z_{2}}=\bar{z_{1}\bar{z_{2}}}$
• $\left ( \bar{\frac{z_{1}}{z_{2}}} \right )=\frac{\bar{z_{1}}}{\bar{z_{2}}}$ provided $z_{2}$ ≠ 0.

Example: $\frac{5+\sqrt{2i}}{1-\sqrt{2i}}$

Solution: $\frac{5+\sqrt{2i}}{1-\sqrt{2i}}=\frac{5+\sqrt{2i}}{1-\sqrt{2i}}\times \frac{1+\sqrt{2i}}{1+\sqrt{2i}}=\frac{5+5\sqrt{2i}+\sqrt{2i-2}}{1-(\sqrt{2i}^{}2)}$

We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where discriminant is non-negative, i.e., ≥ 0,
Let us consider the following quadratic equation: $ax^{2}+bx+c\: =\: 0$ with real coefficients a, b, c and a ≠ 0. Also, let us assume that the $b^{2}-4ac\: < 0$
Now, we know that we can find the square root of negative real numbers in the set of complex numbers. Therefore, the solutions to the above equation are available in the set of complex numbers which are given by
x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\:=\:\frac{-b\pm \sqrt{4ac-b^{2}}i}{2a}$

Note: At this point of time, some would be interested to know as to how many roots does an equation have? In this regard, the following theorem known as the Fundamental theorem of Algebra is stated below (without proof).

“A polynomial equation has at least one root.”
As a consequence of this theorem, the following result, which is of immense importance, arrived at
“A polynomial equation of degree n has n roots.”

Example: solve $x^{2}+2\: =\: 0$

Solution: $x^{2}\: =\: -2\: i.e\: x\: =\: \pm \sqrt{-2}\: =\: \pm \sqrt{2}i$

Example: solve $x^{2}+x+1\: =\: 0$

Solution: $b^{2}-4ac=1^{2}- 4\times 1\times 1$ = 1-4 = -3

Point to be Remember:

• A number of the form a + ib, where a and b are real numbers, is called a complex number, a is called the real part and b is called the imaginary part of the complex number
• $z_{1}$ = a + ib and $z_{2}$ = c + id . Then
(i) $z_{1}+ z_{2}$ = (a + c) + i (b + d)
(ii) $z_{1} z_{2}$ = (ac – bd) + i (ad + bc)
• For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists the complex number
$\frac{a}{a^{2}+b^{2}}+i\frac{-b}{a^{2}+b^{2}}$ , denoted by $\frac{1}{z}\: orz\: ^{-1}$ called the multiplicative inverse of z such that (a + ib) $\frac{a^{2}}{a^{2}+b^{2}}+i\frac{-b}{a^{2}+b^{2}}$ = 1 + i0 =1
• For any integer k, $i^{4k}\: =\: 1,i^{4k+1}=\: \: i,i^{4k+2}\: =\: -1,i^{4k+3}\: =\: -i$
• The conjugate of the complex number z = a + ib, denoted by $\bar{z}$ , is given by by $\bar{z}$ = a + ib
• The polar form of the complex number z = x + iy is r (cosθ + i sinθ), where $r=\sqrt{x^{2}+y^{2}}$ ( the module of z) and $cos\Theta =\frac{x}{r},sin\Theta =\frac{y}{r}$ (θ is known as the argument of z. The value of θ, such that – π < θ ≤ π, is called the principal argument of z
• A polynomial equation of n degree has n roots.
• The solutions of the quadratic equation $ax^{2}+bx+c=0$ where a, b, c ∈ R, a ≠ 0, b2 – 4ac < 0, are given by x = $\frac{-b\pm \sqrt{4ac-b^{2}}i}{2a}$

## Important Videos on Complex Number and Quadratic Equations - Must Watch

Number Sets

Number Sets 1

Number Sets 2

Number Sets 3

Introduction to i and Imaginary Numbers

Calculating i Raised to Arbitrary Exponents

Complex Numbers

Complex Numbers Part 1

Complex Numbers Part 2

i as the Principal Root of -1 (a little technical)

IIT JEE Complex Numbers (part 1)

IIT JEE Complex Numbers (part 2)

IIT JEE Complex Numbers (part 3)

Algebra II: Imaginary and Complex Numbers

Imaginary Roots of Negative Numbers

Complex Conjugates Example

Subtracting Complex Numbers

Multiplying Complex Numbers

Dividing Complex Numbers

Complex Roots from the Quadratic Formula

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