Solving Complex Number Equations
This article will guide you through solving the complex number equation (x + iy)(2 - 3i) = 4 + i, where x and y are real numbers.
Understanding Complex Numbers
Complex numbers are numbers of the form a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit, defined as the square root of -1 (i² = -1).
Expanding the Equation
First, we need to expand the left side of the equation using the distributive property (FOIL method):
(x + iy)(2 - 3i) = 2x - 3xi + 2yi - 3yi²
Since i² = -1, we can simplify this to:
2x - 3xi + 2yi + 3y = (2x + 3y) + (-3x + 2y)i
Equating Real and Imaginary Components
Now, we have the equation:
(2x + 3y) + (-3x + 2y)i = 4 + i
For two complex numbers to be equal, their real and imaginary components must be equal. This gives us two separate equations:
- 2x + 3y = 4
- -3x + 2y = 1
Solving the System of Equations
We can solve this system of linear equations using various methods, like substitution or elimination. Here, we will use elimination:
- Multiply the first equation by 3 and the second equation by 2:
- 6x + 9y = 12
- -6x + 4y = 2
- Add the two equations together:
- 13y = 14
- Solve for y:
- y = 14/13
- Substitute the value of y back into either of the original equations to solve for x. Let's use the first equation:
- 2x + 3(14/13) = 4
- 2x = 4 - 42/13
- 2x = 10/13
- x = 5/13
Solution
Therefore, the solution to the equation (x + iy)(2 - 3i) = 4 + i is:
x = 5/13 and y = 14/13