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 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

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

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 modulus of complex numbers are as follows:

The identities of arguments are as follows:

## Conjugate Of A Complex Number

A number consisting of equal real and imaginary part which is equal in magnitude but with opoosite sign can be termed as a complex comjugate of a complex number.

The properties of the conjugate of complex numbers are given by:

Let z and w be two complex numbers.

## Algebra Of Complex Numbers

The different rules for operations on complex numbers 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 center 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 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