ADD AND SUBTRACT COMPLEX NUMBERS
Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
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
Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
About Us | Contact Us | Privacy Policy
©All rights reserved. onlinemath4all.com
Recent Articles
-
Solving Exponential Equations
Feb 23, 26 10:06 AM
Solving Exponential Equations - Concept - Examples -
SAT Math Problems on Exponents and Radicals
Feb 21, 26 08:28 PM
SAT Math Problems on Exponents and Radicals -
SAT Math Problems on Angles
Feb 21, 26 08:20 PM
SAT Math Problems on Angles
