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High School Mathematics - 22.4 Subtraction and Division of Complex Numbers

Subtraction
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
_
z

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

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

 Q 1: (-2-i5) /(3-i6)9/15 i2+i8/15 - 9/15i Q 2: (5+i9)/ (-3+i4)7i21/25 - 47/25i-4+0i Q 3: Write the complex number z = (2+i)/(1+i)(1-i2) in the x+iy form.4/3 - i/31/2 + i/23i/4 Q 4: (1-i)-(-1+i6)2+i52-i71-i Q 5: z1 = 3+i, z2 = 1+i, find the quotient z2/z1i/52/52/5 + i/5 Q 6: Find the sum of the complex number -Ö3+Ö-2 and 2Ö3-i.Ö-3+iÖ3+i(Ö2-1)Ö2+i Q 7: Find the difference of the complex numbers z1> = -3+i2, z2 = 13-i16+3i19-3i-16+i3 Q 8: [(Ö5+i/2) (Ö5-i2)]/6+i5)0+i(72-15Ö5)/122 - (30+9 Ö5)/61 i4/25+i3/5 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!

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