**Complex numbers** are algebraic expressions which have real and imaginary parts. If the real part of a complex number is 0, then it is called a βpurely imaginary numberβ. This article gives insight into complex numbers definition and complex numbers solved examples for aspirants so that they can start with their preparation.

## Complex Number Definition

A number of the form **z = x + iy** where x is the real part, y is the imaginary part and x, y belong to a set of real numbers.Β

## Representation of a Complex Number

The polar form of z : x = r cos ΞΈ, y = r sin ΞΈ

The exponential form of z : z = r e^{iΞΈ} Β (where e^{iΞΈ} = cos ΞΈ + i sin ΞΈ)

## Modulus and Argument of a Complex Number

If z = x + iy, then the modulus is denoted by

The properties of the modulus of complex numbers are as follows:

The identities of arguments are as follows:

## Conjugate of a Complex Number

A number consisting of an equally real and imaginary part which is equal in magnitude but with opposite signs can be termed a complex conjugate of a complex number.

The properties of the conjugate of complex numbers are as follows:

Let z and w be two complex numbers.

## Algebra of Complex Numbers

The different rules for operations on complex numbers are as follows:

## Equality of Complex Numbers

Consider two complex numbers z_{1Β }= (a+bi) and z_{2Β }= (c + di). They are said to be equal if their real and imaginary parts are equal, that is

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## Complex Number Examples

Solved examples on complex numbers are given below.

**Example 1:** If z_{1}, z_{2}, z_{3} are the vertices of an equilateral triangle ABC, such that |z_{1Β }β i| = |z_{2Β }β i| = |z_{3Β }β i|, then what is the value of |z_{1Β }+z_{2Β }+ z_{3}|?

**Solution:**

|z_{1Β }β i| = |z_{2Β }β i| = |z_{3Β }β i|

Hence, z_{1}, z_{2}, and z_{3} lie on the circle whose centre is i.

Also, the circumcenter coincides.

[z_{1 }+ z

_{2 }+ z

_{3}] / 3 = i

β |z_{1 }+ z_{2 }+ z_{3}| = 3

**Example 2: **What is the value of Ξ» if the curve y = (Ξ» + 1)x^{2 }+ 2 intersects the curve y = Ξ»x + 3 at exactly one point?

**Solution: **

As (Ξ» + 1)x^{2 }+ 2 = Ξ»x + 3 has only one solution, so D = 0

β Ξ»^{2 }β 4(Ξ» + 1)(β1) = 0

Or

Ξ^{2 }+ 4Ξ» + 4 = 0

Or (Ξ» + 2)^{2 }= 0

Therefore, Ξ» = β2

**Example 3: **If k + β£k + z^{2}β£ = |z|^{2 };(k β R^{β}), what is the possible argument of z?

**Solution: **

|k + z^{2}| = |z^{2}| β k = |z^{2}| + |k|

β k, z^{2} and 0 + i0 are collinear

β arg (z^{2})= arg (k)

β 2 arg (z) = Ο

β arg (z) = Ο/2

**Example 4: **Let aβ 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and ?a when divided respectively, by x + a and x – a, the remainder when p(x) is divided by x^{2 }β a^{2} is?

**Solution:**

We are given that p(-a) = a and p(a) = -a

(When a polynomial f(x) is divided by x – a, remainder is f[a]),

Let the remainder, when p(x) is divided by x^{2 }β a^{2}, be Ax+B.

Then,

p(x) = Q(x)(x^{2 }β a^{2}) + Ax + B β¦.. (1)

Where Q(x) is the quotient.

Putting x = a and -a in (1), we get

p(a) = 0 + Aa + B

β βa = Aa + B β¦. (2)

And p(βa) = 0 β aA + B

β a = βaA + B β¦β¦.(3)

Solving (2) and (3), we get

B = 0 and A = -1

Hence, the required remainder is -x.

**Example 5: **If the roots of the equation x^{2}+2ax+b=0 are real and distinct and they differ by at most 2m, then at what interval does b lie?

**Solution: **

Let the roots be Ξ±, Ξ².

β΄Ξ± + Ξ² = β2a and Ξ±Ξ² = b

Given, |Ξ± β Ξ²| β€ 2m

or |Ξ± β Ξ²|^{2 }β€ (2m)^{2} or(Ξ± + Ξ²)^{2 }β 4ab β€ 4m^{2} or 4a^{2 }β 4b β€ 4m^{2}

β a^{2 }β m^{2 }β€ b and discriminant D > 0 or 4a^{2 }β 4b > 0

β a^{2 }β m^{2 }β€ b and b < a^{2}.

Hence, b β [a^{2}βm^{2},a^{2}).

**Example 6:** Find the conjugate of (2 – i)(1 + 2i)/(2 + 3i)(3 – 2i).

**Solution:**

We have (2-i)(1+2i)/(2+3i)(3-2i) = (2-i+4i+2)/ (6+9i-4i+6)

= (4+3i) / (12+5i)

Multiply the numerator and denominator by (12-5i).

(4+3i)(12-5i) / (12+5i)(12-5i) = (48+36i-20i+15)/(144+25)

= (63+16i) / 169

Hence, the conjugate is (63+16i) / 169

**Example 7:** Solve the equation x^{2}+3x+9 = 0

**Solution:**

We have x^{2}+3x+9 = 0

b^{2}-4ac = 3^{2}-4Γ1Γ9 = 9-36 = -27 < 0

β΄ x = (-3+β-27)/2 orΒ (-3-β-27)/2Β (Using equation [-bΒ±β(b^{2}-4ac)]/2a)

= (-3+β27 i)/2 orΒ (-3-β27 i)/2Β

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