复数的幂

Powers of Imaginary and Complex Numbers

You can perform any mathematical operation with imaginary and complex numbers! Just like we can add, subtract, multiply, and divide these numbers, we can raise them to powers! Let's first look at raising $i$ and it's multiples to powers. These are the powers of the imaginary unit. $$\begin{align}i^0&=1\\ i^1&=i\\ i^2&=-1\\ i^3&=-i\\ i^4&=1 \end{align}$$ Notice that the even powers result in real numbers and odd powers give us imaginary numbers. There is a repeating pattern of $1, i, -1, -i$ when taking powers of $i$, starting with $i^0$.

If we're trying to simplify $i$ taken to higher powers, we can break down the powers into smaller parts. Remember that $i^4=1$. This means that when we factor out $i^4$, we are just taking out a $1$. We can do this repeatedly until we only have $i^n$, where $n<3$. Let's look at an example, simplifying $i^{58}$. $$\begin{align}i^{58}&=i^{56}\cdot i^2\\ i^{58}&=(i^4)^{14} \cdot i^2 \phantom{00} \text{ Remember } i^4=1\\ i^{58}&={\color{green}{1^{14}}} \cdot i^2\\ i^{58}&=1 \cdot i^2\\ i^{58}&=i^2\\ i^{58}&= -1 \end{align}$$

We can do this for any power. Try this one for yourself:

We can pull out an $i^3$ to make this divisible by $i^4$. Let's take a look. $$\begin{align}i^{91}&=i^{88}\cdot i^{3}\\ i^{91}&=(i^4)^{22} \cdot i^3\\ i^{91}&={\color{green}{1^{22}}} \cdot i^3\\ i^{91}&=1 \cdot i^3\\ i^{91}&=i^3\\ i^{91}&= -i \end{align}$$

Complex Powers

Now let's take a look at raising complex numbers to powers. Raising a complex number, $a+bi$, to a power uses the same process as FOILing polynomials. This is because $(a+bi)^2=(a+bi)(a+bi)$. Let's take a look at this process using the complex number $z=2-3i$. $$\begin{align} (2-3i)^2&=(2-3i)(2-3i)\\ (2-3i)^2&=4-6i-6i+9i^2\\ (2-3i)^2&=4-12i-9\\ (2-3i)^2&=-5-12i \end{align}$$ Let's say we wanted to find $(2-3i)^3$. We need to find $(2-3i)^2$, then multiply it by another $(2-3i)$. $$\begin{align}(2-3i)^3&=(2-3i)(2-3i)(2-3i)\\ (2-3i)^3&=(4-6i-6i+9i^2)(2-3i)\\ (2-3i)^3&=(4-12i-9)(2-3i)\\ (2-3i)^3&=(-5-12i)(2-3i)\\ (2-3i)^3&=-10+15i-24i+36i^2\\ (2-3i)^3&=-10-9i-36\\ (2-3i)^3&=-46-9i \end{align}$$

Try another practice problem to check your understanding.

This one might have less steps than you think! $$\begin{align} (3+i)^4&=((3+i)^2)^2\\ (3+i)^4&=((3+i)(3+i))^2\\ (3+i)^4&=(9+6i+i^2)^2\\ (3+i)^4&=(8+6i)^2\\ (3+i)^4&=(8+6i)(8+6i)\\ (3+i)^4&=64+96i+36i^2\\ (3+i)^4&=28+96i \end{align}$$

We needed to use the FOIL method twice for this problem.

Wrap Up

There are two big points to take away from this section.

1. There is a pattern when taking $i^n$. Starting with $i^0$, the pattern is $1, i, -1, -i$.
2. Taking $(a+bi)^n$ uses the FOIL method, similar to $(x+a)^2$.

(Video) Complex Numbers - Powers of i | Don't Memorise

by Don't Memorise

In this video by Don't Memorise, we can see the derivations of $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$.

We can also learn how to use these values of $i$ to solve for $i$ with much larger exponents.

Using the example of $i^7$ given in the video,

$$i^7 = i^4 * i^2 * i$$ $$i^7 = 1 * -1 * i$$ $$i^7 = -i$$

Don't be afraid to use this method of breaking down the larger exponent into smaller exponents you already know to help you solve the problem.

Powers of Imaginary Numbers

Powers of imaginary numbers follow a pattern that repeats every four powers. Imaginary numbers are defined using square roots of negative real numbers.

The imaginary unit, $i$, is defined as the square root of $-1$, so $i² = -1$. If you keep raising $i$ to higher powers, you'll notice a cycle: $i³ = -i$, $i⁴ = 1$, and then it starts over again. So, $i⁵ = i$, $i⁶ = -1$, $i⁷ = -i$, and $i⁸ = 1$, and this pattern continues forever! You can use this cycle to simplify large powers of $i$.