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# Complex Numbers Solved Examples

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purely 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

$$\begin{array}{l}\text{The value of i} = \sqrt{-1}.\end{array}$$
The value of i can’t be seen on the number line.

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

## Modulus and Argument of a Complex Number

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

$$\begin{array}{l}|z|=\sqrt{a^2+b^2}.\end{array}$$
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:

$$\begin{array}{l}1]\ |z_{1} z_{2}| = |z_{1}\cdot z_{2}|\\ 2]\ |z_{1} + z_{2}| \leq |z_{1} + z_{2}|\\ 3]\ |\frac{z_{1}}{z_{2}}|=\frac{|z_{1}|}{|z_{2}|}\\ 4]\ |z_{1} – z_{2}| \geq |z_{1} – z_{2}|\\\end{array}$$

The identities of arguments are as follows:

$$\begin{array}{l}1] \ \ {Arg}(z_{1}z_{2})\equiv \ {Arg}(z_{1})+\ {Arg}(z_{2}){\pmod {(-\pi ,\pi ]}},\\ 2] \ {\displaystyle \ {Arg} {\biggl (}{\frac {z_{1}}{z_{2}}}{\biggr )}\equiv \ {Arg} (z_{1})-\ {Arg} (z_{2}){\pmod {(-\pi ,\pi ]}}.}\\ 3] \text{If} \ z \neq 0 \ and \ n \ is \ any \ integer, \ then \ {Arg}\left(z^{n}\right)\equiv n\ {Arg}(z){\pmod {(-\pi ,\pi ]}}.\end{array}$$

## 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.

$$\begin{array}{l}1]\ {\overline {z+w}}={\overline {z}}+{\overline {w}}\\ 2]\ {\overline {z-w}}={\overline {z}}-{\overline {w}}\\ 3]\ {\overline {zw}}={\overline {z}}\;{\overline {w}}\\ 4]\ {\overline {\left({\frac {z}{w}}\right)}}={\frac {\overline {z}}{\overline {w}}},\quad {\text{if }}w\neq 0\\ 5]\ {\overline {z}}=z~\Leftrightarrow ~z\in \mathbb {R}\\ 6]\ {\overline {z^{n}}}=\left({\overline {z}}\right)^{n},\quad \forall n\in \mathbb {Z} \\ 7]\ \left|{\overline {z}}\right|=\left|z\right|\\ 8]\ {\left|z\right|}^{2}=z{\overline {z}}={\overline {z}}z\\ 9]\ {\overline {\overline {z}}}=z\\ 10] \ z^{-1}={\frac {\overline {z}}{{\left|z\right|}^{2}}},\quad \forall z\neq 0\end{array}$$

## Algebra of Complex Numbers

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

$$\begin{array}{l}1] \ {\displaystyle z_{1}+z_{2}=(a+bi)+(c+di)=(a+c)+(b+d)i\,}\\ 2] \ {\displaystyle z_{1}-z_{2}=(a+bi)-(c+di)=(a-c)+(b-d)i\,}\\ 3] \ {\displaystyle z_{1}z_{2}=(a+bi)(c+di)=ac+adi+bci+bdi^{2}=(ac-bd)+(bc+ad)i\,}\\ 4] \ \frac{z_{1}}{z_{2}}=\ {\displaystyle{\frac {ac+bd}{c^{2}+d^{2}}}+{\frac {bc-ad}{c^{2}+d^{2}}}i}.\\\end{array}$$

## Equality of Complex Numbers

Consider two complex numbers z1Β = (a+bi) and z2Β = (c + di). They are said to be equal if their real and imaginary parts are equal, that is

$$\begin{array}{l}z_{1} = z_{2} \rightarrow Re(z_{1}) = Re(z_{2})\ \text{and}\ \ Im (z_{1}) = Im (z_{2}).\end{array}$$

### Related articles

Complex Numbers

Representation of a Complex Number

## 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 |z1Β β i| = |z2Β β i| = |z3Β β i|, then what is the value of |z1Β +z2Β + z3|?

Solution:

|z1Β β i| = |z2Β β i| = |z3Β β 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 β 4(Ξ» + 1)(β1) = 0

Or

Ξ2 + 4Ξ» + 4 = 0

Or (Ξ» + 2)2 = 0

Therefore, Ξ» = β2

Example 3: If k + β£k + z2β£ = |z|2 ;(k β Rβ), what is the possible argument of z?

Solution:

|k + z2| = |z2| β 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 β 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 β a2, be Ax+B.

Then,

p(x) = Q(x)(x2 β a2) + 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 βa = Aa + B β¦. (2)

And p(βa) = 0 β aA + B

β a = βaA + 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 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.

β΄Ξ± + Ξ² = β2a 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€ 4m2 or 4a2 β 4b β€ 4m2

β a2 β m2 β€ b and discriminant D > 0 or 4a2 β 4b > 0

β a2 β m2 β€ b and b < a2.

Hence, b β [a2β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

β΄ x = (-3+β-27)/2 orΒ  (-3-β-27)/2Β  (Using equation [-bΒ±β(b2-4ac)]/2a)

= (-3+β27 i)/2 orΒ  (-3-β27 i)/2Β

## Complex Numbers – Important Questions

Test your Knowledge on Complex Numbers