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COMPLEX NUMBERS
The number in the form of a + ib where a and b are real numbers and i = √-1 is called the complex number The square root of negative number (√-a) is called the imaginary number, for example √-1, √-5, √-16 etc all are imaginary numbers. The complex number has two parts which are real part that contains real number and the second is imaginary part which contains imaginary number, for example the complex number z = a + ib, here “a” is real part and “ib” is imaginary part and they are denoted by Re(z) which means real part of complex number z, and Im(z) which means the imaginary part of complex number z.
Example from the complex number z = 2 + 5i, here Re(z) = 2 and Im(z) = 5i.
INTEGRAL POWER OF IMAGINARY UNIT “i”
This is given by:- i = √-1 then i2 = -1 i3 = i2 x i = -1 x √-1 = -i i4 = i2 x i2 = -1 x -1 = 1 i5 = i4 x i = 1 x √-1 = i i6 = i5 x i = i x √-1 = √-1 x √-1 = -1 etc.
EQUALITY OF COMPLEX NUMBER If z1 = a + ib and z2 = c + id are complex numbers in standard form then a + ib = c + id, if and only if a = c and b = d.
ALGEBRA OF COMPLEX NUMBER These are:- 1. Addition 2. Subtraction 3. Multiplication 4. Division 1. ADDITION OF COMPLEX NUMBERS Consider the complex numbers z1 = a + ib and z2 = c + id z1 + z2 = (a + ib) + (c + id) z1 + z2 = a + ib + c + id z1 + z2 = (a + c) + (ib + id)
Therefore z1 + z2 = (a + c) + i(b + d)
PROPERTIES UNDER ADDITION 1. Commutative If z1 and z2 are two complex numbers then z1 + z2 = z2+ z1
2. Associative If z1 and z2 and z3 are three complex numbers then (z1 + z2) + z3 = z1 + (z2 + z3) 3. Additive identity property If z is the complex number, then z + 0 = 0 + z = z for all z
4. Additive inverse property For any complex number z = a + ib there exists –z = -a + i(-b) such that z + (-z) = (-z) + z = 0
2. SUBTRACTION OF COMPLEX NUMBERS Consider the complex numbers z1 = a + ib and z2 = c + id z1 - z2 = (a + ib) - (c + id) z1 - z2 = a + ib - c - id z1 + z2 = (a - c) + (ib - id)
Therefore z1 - z2 = (a - c) + i(b - d) Example 1: Given that, z1 = 5 – 2i and z2 = 3 + 6i find z1 + z2 Solution: Given: - z1 = 5 – 2i z2 = 3 + 6i
Then z1 + z2 = 5 – 2i + 3 + 6i z1 + z2 = 5 + 3 + 6i – 2i Therefore z1 + z2 = 8 + 4i
Example 2: Find the sum of the following complex numbers, 3 + 2i and -4 + 5i Solution: Given: - z1 = 3 + 2i z2 = -4 + 5i Then z1 + z2 = 3 + 2i - 4 + 5i z1 + z2 = 3 - 4 + 2i + 5i Therefore z1 + z2 = -1 + 7i Example 3: If z1 = 12 + 7i z2 = 9 – 3i z3 = 17 + 4i
verify that, (z1 + z2) + z3 = z1 + (z2 + z3)
Solution: Given z1 = 12 + 7i z2 = 9 – 3i z3 = 17 + 4i then consider Left Hand Side (L.H.S)
(z1 + z2) + z3 = (12 + 7i + 9 – 3i) + 17 + 4i (z1 + z2) + z3 = (12 + 9 + 7i – 3i) + 17 + 4i (z1 + z2) + z3 = 21 + 4i + 17 + 4i (z1 + z2) + z3 = 21 + 17 +4i + 4i Therefore (z1 + z2) + z3 = 38 + 8i
Consider Right Hand Side (R.H.S) z1 + (z2 + z3) = 12 + 7i + (9 – 3i + 17 + 4i) z1 + (z2 + z3) = 12 + 7i + (9 + 17 +4i – 3i) z1 + (z2 + z3) = 12 + 7i + (26 +i) z1 + (z2 + z3) = 12 + 7i + 26 + i = 12 + 26 + 7i + i Therefore z1 + (z2 + z3) = 38 + 8i
Since (z1 + z2) + z3 = z1 + (z2 + z3) = 38 + 8i Hence (z1 + z2) + z3 = z1 + (z2 + z3), Verified. Example 4: Given that, z1 = 3 – 2i and z2 = 5 + 2i find z1 - z2
Solution: Given: - z1 = 3 – 2i z2 = 5 + 2i
Then z1 - z2 = (3 – 2i) – (5 + 2i) z1 - z2 = 3 – 2i – 5 – 2i z1 - z2 = 3 – 5 – 2i – 2i z1 - z2 = -2 – 4i
Therefore z1 - z2 = -2 - 4i
3. MULTIPLICATION OF COMPLEX NUMBERS
Consider the complex numbers:- z1 = a + ib and z2 = c + id
Then
z1 x z2 = (a + ib) x (c + id) z1 x z2 = ac + iad + ibc + i2bd (Remember here i2 = -1) z1 x z2 = ac – bd + i(ad + bc)
Therefore z1 x z2 = ac – bd + i(ad + bc)
Example 5: If z1 = 4 + 3i and z2 = 3 – 2i, find z1 x z2
Solution: z1 x z2 = (4 + 3i) x (3 – 2i) z1 x z2 = 12 – 8i + 9i – 6i2 (Remember here i2 = -1) z1 x z2 = 12 + 6 + i = 18 + i
Therefore z1 x z2 = 18 + i
PROPERTIES UNDER ADDITION
1. Commutative If z1 and z2 are two complex numbers then z1 x z2 = z2 x z1
2. Associative If z1 and z2 and z3 are three complex numbers then (z1 x z2) x z3 = z1 x (z2 x z3)
3. Multiplicative identity of complex number If z is any complex number then z x 1 = 1 x z = z
3. Multiplicative inverse of complex number For every non- zero complex number z = a + ib, there exists a complex number z1 = x + iy such that z x z1 = z1 x z = 1.
(x + iy)(a + ib)=1 x (a + ib) + iy (a + ib) = 1 ax + ibx +iay + i2by = 1 ax + ibx + iay – by = 1 (i2by became –by since i2 = -1) ax – by + i(bx + ay) = 1 + 0i ( we added 0i in order to make identity both sides)
by equating real and imaginary parts of complex number:-
ax – by = 1 ……………………(i) bx + ay = 0 ……………………(ii)
By solving equations (i) and (ii) simultaneously we obtain:-
X = a/(a2 + b2) and y = -b/(a2 + b2)
From z1 = x + iy we substitute values we obtained above in this equation.
Z1 = a/(a2 + b2) + i (-b/(a2 + b2))
Therefore For every non – zero complex number z = a + ib, multiplicative inverse is given by a/(a2 + b2) + i (-b/(a2 + b2)) CONJUGATE OF COMPLEX NUMBER
A pair of complex number z1 and z2 is said to be conjugate of each other if the sum and product of two complex numbers are both real.
Let z1 = a + ib and z2 = a – ib Here the sum = a + ib + a – ib = 2a which is real.
Also the product = (a + ib) x (a – ib) = a2 – iab + iab – i2b2 = a2 + b2 which is real.
The conjugate of complex number z is obtained by changing the sign of imaginary part of z and it is denoted by Z̅.
Example, given that z = a + ib, find z̅. Solution, z̅ = a – ib. *************** THANKS FOR READING ******** | |
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