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High School Mathematics - 2
2.4 Subtraction and Division of Complex Numbers

We know that for any complex numbers z1 and z2, there exists a complex number z such that z1 + z = z2. This number z is denoted by z2-z1. Let z1 = a+ib, z2 = c+id and z = x+iy, then
z1 + z = z2 or (a+ib) + (x+iy) = c+id
i.e (a+x) + i(b+y) = c+id
a+x = c, b+y = d
This system of equations has a unique solution
x = c-a, y = d-b
Thus z = (c-a) + i(d-b)
Consequently, the difference z2 - z1 always exist with
z = z2 - z1 = (c+id) - (a+ib) = (c-a) + i(d-b) which yields the rule of subtraction of complex numbers.

Division of Complex Numbers
We know that for any complex numbers z1 and z2 (not equal to 0), there exists a complex number z such that
z1 = zz2 ----1 The number z , is denoted by z1/z2 called the division of complex numbers
Let us consider two complex numbers z1 = a+ib, z2 = c+id
z1/z2 = (a+ib)/(c+id)x(c-id)/(c-id) multiply the numerator and denominator by
(ac+bd)/(c2+d2) + (bc-ad)i/(c2+d2)

Example: Given the complex numbers z1> = 3+i, z2 = 1+i, find the quotient z2/z1
Solution: z2/z1 = (1+i)/(3+i)
(1+i)/(3+i) x (3-i)/(3-i) = (4+i2)/(9+1) = 2/5 + i/5
Answer: 2/5 + i/5

Directions: Solve the following questions. Also write at least 5 examples of your own for subtaction and division of complex numbers.

Q 1: (7-i2)-(4+i)+(-3+i5)

Q 2: Find the sum of the complex number -3+-2 and 23-i.

Q 3: (-2-i5) /(3-i6)
9/15 i
8/15 - 9/15i

Q 4: Write the complex number z = (2+i)/(1+i)(1-i2) in the x+iy form.
1/2 + i/2
4/3 - i/3

Q 5: [(5+i/2) (5-i2)]/6+i5)
(72-155)/122 - (30+9 5)/61 i

Q 6: Find the difference of the complex numbers z1> = -3+i2, z2 = 13-i

Q 7: (5+i9)/ (-3+i4)
21/25 - 47/25i

Q 8: (1-i)-(-1+i6)

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!

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