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 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
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 z1Β = (a+bi) and z2Β = (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 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 = iβ |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)(β1) = 0
Or
Ξ2 + 4Ξ» + 4 = 0
Or (Ξ» + 2)2 = 0
Therefore, Ξ» = β2
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 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 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 = 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€ 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 [a2βm2,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 = 32-4Γ1Γ9 = 9-36 = -27 < 0
β΄ x = (-3+β-27)/2 orΒ (-3-β-27)/2Β (Using equation [-bΒ±β(b2-4ac)]/2a)
= (-3+β27 i)/2 orΒ (-3-β27 i)/2Β
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