ADD AND SUBTRACT COMPLEX NUMBERS
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The number which is in the form of a + ib is known as complex number. Every complex numbers will have two parts. They are real part and imaginary part.
To add and subtract complex numbers, we have to combine the real parts together and imaginary parts together.
(a + ib) + (c + id) :
= (a + c) + (ib + id)
= (a + c) + i(c + d)
Example 1 :
Add (-3i) and (3 + 5i).
Solution :
(-3i) + (3 + 5i) = 3 + (-3i + 5i)
= 3 + 2i
Example 2 :
Add (-6 - 2i) and (6 - 5i).
Solution :
(-6 - 2i) + (6 - 5i) = (-6 + 6) + (-2i - 5i)
= 0 + (-7i)
= -3i
Example 3 :
Add (5 + 6i) and (2 - 7i).
Solution :
(5 + 6i) + (2 - 7i) = (5 + 2) + (6i - 7i)
= 7 + (-i)
= 7 - i
Example 4 :
Add (5 - 6i), 5i and (7 + 6i).
Solution :
(5 - 6i) + 5i + (7 + 6i) = (5 + 7) + (-6i + 5i + 6i)
= 12 + 5i
Example 5 :
Simplify : (-7 + 7i) - (-7 - 3i) + (-7 - 8i).
Solution :
(-7 + 7i) - (-7 - 3i) + (-7 - 8i) = -7 + 7i + 7 + 3i - 7 - 8i
= (-7 + 7 - 7) + (7i + 3i - 8i)
= -7 + 2i
Example 6 :
Simplify : (-4 - 7i) - (4 + 5i) - (2 - i).
Solution :
(-4 - 7i) - (4 + 5i) - (2 - i) = -4 - 7i - 4 - 5i - 2 + i
= (-4 - 4 - 2) + (-7i - 5i + i)
= -10 + (-11i)
= -10 - 11i
Example 7 :
Simplify : (1 + 6i) + (6 - 2i) - (-7 + 5i).
Solution :
(1 + 6i) + (6 - 2i) - (-7 + 5i) = 1 + 6i + 6 - 2i + 7 - 5i
= (1 + 6 + 7) + (6i - 2i - 5i)
= 14 + (-i)
= 14 - i
Example 8 :
Simplify : (-5 + 7i) - (-6 + i) - (-6 + 5i).
Solution :
(-5 + 7i) - (-6 + i) - (-6 + 5i) = -5 + 7i + 6 - i + 6 - 5i
= (-5 + 6 + 6) + (7i - i - 5i)
= 7 + i
Example 9 :
Subtract (3 - 4i) from (8 + 2i).
Solution :
(8 + 2i) - (3 - 4i) = 8 + 2i - 3 + 4i
= (8 - 3) + (2i + 4i)
= 5 + 6i
Example 10 :
Subtract (-5 - i) from (2 - 7i).
Solution :
(2 - 7i) - (-5 - i) = 2 - 7i + 5 + i
= (2 + 5) + (-7i + i)
= 7 + (-6i)
= 7 - 6i
Example 11 :
(1 + i)2 + (1 - i)2
Solution :
= (1 + i)2 + (1 - i)2
|
(1 + i)2 = 12 + 2(1)(i) + i2 = 1 + 2i - 1 = 2i |
(1 - i)2 = 12 - 2(1)(i) + i2 = 1 - 2i - 1 = -2i |
(1 + i)2 + (1 - i)2 = 2i - 2i
= 0
Example 12 :
(1 + i)8 + (1 - i)8
Solution :
= (1 + i)8 + (1 - i)8
|
(1 + i)8 = [(1 + i)2]4 = [12 + 2(1)(i) + i2]4 = (1 + 2i - 1)4 = (2i)4 = 16i4 = 16(i2)2 = 16(-1)2 = 16 |
(1 - i)8 = [(1 - i)2]4 = [12 - 2(1)(i) + i2]4 = (1 - 2i - 1)4 = (-2i)4 = 16i4 = 16(i2)2 = 16(-1)2 = 16 |
(1 + i)8 + (1 - i)8 = 16 + 16
= 32
So, the answer is 32.
Find the values of x and y that make each equation true.
Example 13 :
9 + 12i = 3x + 4iy
Solution :
Given that, 9 + 12i = 3x + 4iy
Since we have complex numbers on both sides of the equal sign, we have to equate the real parts and imaginary parts.
9 = 3x and 12 = 4y
x = 9/3 and y = 12/4
x = 3 and y = 3
Example 14 :
x + 1 + 2yi = 3 - 6i
Solution :
Given that, x + 1 + 2yi = 3 - 6i
(x + 1) + i (2y) = 3 - 6i
Equating the real and imaginary parts.
x + 1 = 3 and 2y = -6
x = 3 - 1 and y = -6/2
x = 2 and y = -3
Example 15 :
(2x + 7) + (3 - y)i = -4 + 6i
Solution :
(2x + 7) + (3 - y)i = -4 + 6i
2x + 7 = -4 and 3 - y = 6
2x = -4 - 7 and y = 3 - 6
2x = -11 and y = -3
x = -11/2 and y = -3
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