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 β€œpurely imaginary number”. 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. 

The value of i=βˆ’1.
The value of i can’t be seen on the number line.

Representation of a Complex Number

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 eiΞΈ  (where eiΞΈ = cos ΞΈ + i sin ΞΈ)

Modulus and Argument of a Complex Number

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

|z|=a2+b2.
The angle made by the positive direction of the real axis can be defined as the argument of a complex number and denoted by arg(z).

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

1] |z1z2|=|z1β‹…z2|2] |z1+z2|≀|z1+z2|3] |z1z2|=|z1||z2|4] |z1–z2|β‰₯|z1–z2|

The identities of arguments are as follows:

1]  Arg(z1z2)≑ Arg(z1)+ Arg(z2)(mod(βˆ’Ο€,Ο€]),2]  Arg(z1z2)≑ Arg(z1)βˆ’ Arg(z2)(mod(βˆ’Ο€,Ο€]).3]If zβ‰ 0 and n is any integer, then Arg(zn)≑n Arg(z)(mod(βˆ’Ο€,Ο€]).

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.

1] z+w―=z―+w―2] zβˆ’w―=zβ€•βˆ’w―3] zw―=z―w―4] (zw)―=z―w―,if wβ‰ 05] z―=z β‡” z∈R6] zn―=(z―)n,βˆ€n∈Z7] |z―|=|z|8] |z|2=zz―=z―z9] z――=z10] zβˆ’1=z―|z|2,βˆ€zβ‰ 0

Algebra of Complex Numbers

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

1] z1+z2=(a+bi)+(c+di)=(a+c)+(b+d)i2] z1βˆ’z2=(a+bi)βˆ’(c+di)=(aβˆ’c)+(bβˆ’d)i3] z1z2=(a+bi)(c+di)=ac+adi+bci+bdi2=(acβˆ’bd)+(bc+ad)i4] z1z2= ac+bdc2+d2+bcβˆ’adc2+d2i.

Equality of Complex Numbers

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

z1=z2β†’Re(z1)=Re(z2) and  Im(z1)=Im(z2).

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Complex Number Examples

Solved examples on complex numbers are given below.

Example 1: If z1, z2, z3 are the vertices of an equilateral triangle ABC, such that |zβˆ’ i| = |zβˆ’ i| = |zβˆ’ i|, then what is the value of |z+z+ z3|?

Solution:

|zβˆ’ i| = |zβˆ’ i| = |zβˆ’ i|

Hence, z1, z2, and z3 lie on the circle whose centre is i.

Also, the circumcenter coincides.

[z1 + z2 + z3] / 3 = i

β‡’ |z1 + z2 + z3| = 3

Example 2: What is the value of Ξ» if the curve y = (Ξ» + 1)x2 + 2 intersects the curve y = Ξ»x + 3 at exactly one point?

Solution:

As (Ξ» + 1)x2 + 2 = Ξ»x + 3 has only one solution, so D = 0

β‡’ Ξ»2 βˆ’ 4(Ξ» + 1)(βˆ’1) = 0

Or

Ξ›2 + 4Ξ» + 4 = 0

Or (Ξ» + 2)2 = 0

Therefore, Ξ» = βˆ’2

Example 3: If k + ∣k + z2∣ = |z|2 ;(k ∈ Rβˆ’), what is the possible argument of z?

Solution:

|k + z2| = |z2| βˆ’ k = |z2| + |k|

β‡’ k, z2 and 0 + i0 are collinear

β‡’ arg (z2)= 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 x2 βˆ’ a2 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 x2 βˆ’ a2, be Ax+B.

Then,

p(x) = Q(x)(x2 βˆ’ a2) + Ax + B ….. (1)

Where Q(x) is the quotient.

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

p(a) = 0 + Aa + B

β‡’ βˆ’a = Aa + B …. (2)

And p(βˆ’a) = 0 βˆ’ aA + B

β‡’ a = βˆ’aA + B …….(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 x2+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.

∴α + Ξ² = βˆ’2a and Ξ±Ξ² = b

Given, |Ξ± βˆ’ Ξ²| ≀ 2m

or |Ξ± βˆ’ Ξ²|2 ≀ (2m)2 or(Ξ± + Ξ²)2 βˆ’ 4ab ≀ 4m2 or 4a2 βˆ’ 4b ≀ 4m2

β‡’ a2 βˆ’ m2 ≀ b and discriminant D > 0 or 4a2 βˆ’ 4b > 0

β‡’ a2 βˆ’ m2 ≀ b and b < a2.

Hence, b ∈ [a2βˆ’m2,a2).

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 x2+3x+9 = 0

Solution:

We have x2+3x+9 = 0

b2-4ac = 32-4Γ—1Γ—9 = 9-36 = -27 < 0

∴ x = (-3+√-27)/2 or  (-3-√-27)/2  (Using equation [-b±√(b2-4ac)]/2a)

= (-3+√27 i)/2 or  (-3-√27 i)/2 

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