Imaginary numbers

Question

Find the values of x and y that satisfy the equation 14x - 64i = 42 + yi

Answer 1

As far as algebra is concerned, imaginary numbers can be worked with pretty much just like real numbers. All the usual rules of algebra apply.

So we have an equation:

14x - 64i = 42 + yi

To get this in a more standard form, we want to write y as a function of x. So we do the following operations:

1. Flip the equality to get: 42 + yi = 14x - 64i
2. Subtract 42 from both sides to get: y i = 14 x -64 i - 42
3. Divide both sides by i to get: y = 14 x/i - 64 - 42/i
4. Here's where a fact about i comes in: Since i * i = -1, we can write i = -1/i. Multiplying both sides by -1 gives: -i = 1/i. So we can replace 1/i by -i. Dividing by i is the same as multiplying by 1/i, which is the same as multiplying by -i. So the equation in 3 can be rewritten as: y = -14 x i - 64 + 42 i.

So the solutions to the equation are found by any pair of numbers (x,y) where y = -14 x i - 64 + 42 i. For example, we can let x=0 to find that one solution is y = -64 + 42 i. So the pair(0, -64 + 42 i) is one solution. We can check by plugging those into the original equation:

14x - 64i = 42 + yi

becomes

14*0 - 64i = 42 + (-64 + 42 i) * i

Simplifying gives:

-64i = 42 - 64 i + 42 i * i

Since i*i = -1, this simplifies further to:

-64i = 42 - 64i - 42

which is true.

Answer 2

Equate the real parts and imaginary parts:

14x = 42

x = ___?

-64i = yi

___ = y ?