For example, if we have an equation like x^2 + 5x 10 = 0, we can solve it by finding values of x that make this expression equal to zero. But what happens when there are imaginary numbers involved? Let’s take a look at an example:

(x+3i)^2 = 9-4i

This looks pretty weird, but let’s break it down step by step. First, we square the term inside the parentheses using the power rule for exponents (which is just like squaring regular numbers):

(x+3i) * (x+3i) = x^2 + 6xi + 9i^2

Now, let’s deal with that ***** i^2. Remember how we said earlier that multiplying an imaginary number by itself gives us a negative real number? Well, this is where it comes in handy:

-1 * (-1) = 1 (because the product of two negatives is positive)
i * i = -1 (because the square of any imaginary number is always equal to -1)

So when we substitute these values into our equation, we get:

x^2 + 6xi + 9(-1) = x^2 + 6xi 9

Now that we’ve simplified things a bit, let’s take another look at the original expression and see if it matches up with what we just calculated. If they do match up, then we know we’ve found a solution to our equation!

(x+3i)^2 = 9-4i
x^2 + 6xi 9 = 0 (our simplified version)

To solve for x, we can use the quadratic formula:

x = (-b ± sqrt(b^2 4ac)) / 2a

In this case, b is equal to 6i and a and c are both equal to 1. So let’s plug those values into our formula:

x = (-6i ± sqrt((-6i)^2 4(1)(9))) / (2(1))

Simplifying this expression gives us:

x = (-3/2) ± ((sqrt(-54)/2)*i)

We’ve solved a complex equation using the power rule for exponents, some basic algebraic manipulations, and a little bit of imaginary number magic.