How do you perform operations on complex numbers?

Complex Numbers: They’re Not That Complicated (I Promise!)

Okay, complex numbers. I know, the name itself can sound a bit intimidating, right? You see that little i hanging around—the square root of -1—and suddenly things feel…well, complex. But trust me, once you get the hang of it, working with these numbers is surprisingly straightforward, and they pop up everywhere in math, physics, and engineering. So, let’s break it down and make it less scary, shall we?

At their heart, complex numbers are just numbers with two parts: a real part and an imaginary part. Think of them as a + bi, where a and b are your regular, run-of-the-mill real numbers, and i is that imaginary unit we talked about.

Now, let’s get to the good stuff: how to actually use these things.

Adding and Subtracting: It’s Like Combining Apples and Oranges (Sort Of)

Adding and subtracting complex numbers is probably the easiest part. Basically, you just group the “real” stuff together and the “imaginary” stuff together. Simple as that!

• Adding: (a + bi) + (c + di) = (a + c) + (b + d)i
• Subtracting: (a + bi) – (c + di) = (a – c) + (b – d)i

For instance:

(3 + 4i) + (-1 + 2i) = (3 – 1) + (4 + 2)i = 2 + 6i. See? We just added the 3 and the -1, and then the 4i and the 2i.

And for subtraction: (7 – 2i) – (4 + i) = (7 – 4) + (-2 – 1)i = 3 – 3i. Easy peasy.

Think of it like this: you can only add “real” numbers with other “real” numbers, and “imaginary” numbers with other “imaginary” numbers. You wouldn’t add apples and oranges, would you? (Well, maybe in a smoothie, but that’s a different story!)

Multiplication: Time to Distribute!

Multiplying complex numbers is where things get a little more interesting. Remember the distributive property from algebra? (Also sometimes called FOIL – First, Outer, Inner, Last). That’s your best friend here. Also, the golden rule: i2 = -1. Don’t forget that!

Here’s the formula: (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad + bc)i

Let’s do an example:

(2 + 3i)(1 – i) = 2(1) + 2(-i) + 3i(1) + 3i(-i) = 2 – 2i + 3i – 3i2 = 2 + i – 3(-1) = 2 + i + 3 = 5 + i

Notice how we replaced that i2 with -1? That’s the key!

Division: The Conjugate is Your Secret Weapon

Okay, division is where things get slightly tricky, but nothing you can’t handle. The trick is to get rid of that imaginary part in the denominator. How do we do that? By multiplying both the top and bottom of the fraction by the conjugate of the denominator. The conjugate of a + bi is simply a – bi. You just flip the sign in front of the imaginary part.

The formula looks like this: (a + bi) / (c + di) = (a + bi)(c – di) / (c + di)(c – di) = (ac + bd) + (bc – ad)i / (c2 + d2) = (ac + bd)/(c2 + d2) + (bc – ad)/(c2 + d2)i

Yeah, I know, it looks messy. But let’s walk through an example:

(3 + 4i) / (1 – 2i) = (3 + 4i)(1 + 2i) / (1 – 2i)(1 + 2i) = (3 + 6i + 4i + 8i2) / (1 + 4) = (3 + 10i – 8) / 5 = (-5 + 10i) / 5 = -1 + 2i

See? By multiplying by the conjugate, we ended up with a real number in the denominator, making the division possible.

Exponentiation: Polar Form to the Rescue!

Raising a complex number to a power? That’s where polar form becomes your friend. Instead of thinking of a complex number as a + bi, you can think of it in terms of its distance from the origin (r) and the angle it makes with the x-axis (θ).

So, z = a + bi becomes z = r(cos θ + i sin θ), where:

• r = √(a2 + b2)
• θ = arctan(b/a)

Now, here’s where the magic happens: De Moivre’s Theorem. This theorem gives us a shortcut for calculating the power of a complex number in polar form:

zn = rn(cos(nθ) + i sin(nθ))

Example Time: Let’s calculate (1 + i)4

Finding Roots: Digging Deeper with De Moivre

If raising to a power is fun, finding roots is like going on an mathematical treasure hunt! A complex number has n distinct nth roots. To find them, we lean on polar form and De Moivre’s Theorem once again.

The nth roots of z = r(cos θ + i sin θ) are given by:

wk = n√r cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)

where k = 0, 1, 2, …, n-1

Let’s find the cube roots of 8:

So, the cube roots of 8 are 2, -1 + i√3, and -1 – i√3. Pretty cool, huh?

Final Thoughts

Complex numbers might seem a bit weird at first, but they’re incredibly useful and, dare I say, even kind of elegant. Once you understand the basic operations—addition, subtraction, multiplication, division, and how to handle exponents and roots—you’ll be well on your way to unlocking a whole new world of mathematical possibilities. And who knows? You might even start to enjoy them!