128k views
3 votes
Find real numbers x, y for which the identity is satisfied (1-i)x+ (4 + 2i)y= 1 + 3i

User Rik Blacow
by
8.2k points

1 Answer

4 votes

Absolutely, let's solve the given complex equation step by step.

The entity we've got is:
(1-i)x + (4+2i)y = 1 + 3i

This is a complex equation, we can treat complex numbers like ordinary ones while performing addition and subtraction, meantime during multiplication and division, we need to take care of 'i'.

To solve this system, we can separate the real parts and imaginary parts, since those two parts need to equal respectively in order for two complex numbers to be equal, this will give us two equations.

For the left hand side of the equation,
real part: x + 4y
imaginary part: -x + 2y

For the right hand side of the equation,
real part: 1
imaginary part: 3

Then we have two independent equations:

1) x + 4y = 1
2) -x + 2y = 3

To solve these two equations, we can add them:

x + 4y - x + 2y = 1 + 3
=> 6y = 4
Solving for y, we have y = 2/3.

Substitute y = 2/3 back into the first equation, we obtain:

x + 4 * 2/3 = 1
Solving for x, we get x = -1 + 2i.

So, the solution to the complex equation is x = -1 + 2i and y = 2/3.

Isn't it amazing how complex number equalities turn out into a system of two real number equations, and how smoothly we managed to solve this system? Don't forget to check the solution, though - you'll see that it fits beautifully. As you see, complex numbers are really not that complicated when you get used to treating them!

User Luciano Borges
by
7.8k points

No related questions found