By definition, we know that \(i=\sqrt{-1}\) and thus \(i^2=-1\) and we also have that \(i^0=1\) (since any number to the exponent of \(0\) is equal to 1). But what if we wanted to find \(i^{2044}\)?

If we list out the first few positive integral powers of \(i\), we can see a pattern starting to form:

\(i^3=-i,\ i^4=1, \ i^5=i,\ i^6=-1, \ i^7=-i, \ i^8=1, \ i^9=i, \ i^{10}=-1, \ ...\)

A similar pattern for the first few negative integral powers of \(i\) occurs as well:

\(i^{-1}=\frac{1}{i}=-i\),

\(i^{-2}=\frac{1}{i^2}=\frac{1}{-1}=-1\),

\(i^{-3}=i^{-2}\cdot i^{-1}=-1(-i)=i\),

\(i^{-4}=i^{-2}\cdot i^{-2}=-1(-1)=1\), ...

So we can summarize these observations as follows:

And now we have a quick way to determine larger integral powers of \(i\)!

So we can apply this to our previous example:

Example: Evaluate \(i^{2044}\).

Since 2044 is divisible by 4 (i.e. \(4k=2044\) when \(k=511\)), \(i^{2044}=1\).

Now what if we wanted to evaluate integral powers of complex numbers with both a real __and__ imaginary part?

To do so, we first have to introduce **DeMoivre's Formula**, which will allow us to raise complex numbers to any exponent when its in polar form (to review rectangular and polar form conversion see this page):

\((cos(x)+isin(x))^n=cos(nx)+isin(nx)\),

for any integer \(n\) and real number \(x\).

We can apply this to evaluating integral powers of complex numbers (in polar form) as follows:

\((r(cos(\theta)+isin(\theta))^n\)

\(=r^n(cos(\theta)+isin(\theta))^n\)

\(=r^n(cos(n\theta)+isin(n\theta))\)

Example: Evaluate \((-3-3i)^4\).

**Solution:**

1. Notice that \(-3-3i\) is in rectangular form, so we must convert it to polar form before applying the formula. We converted this number as an example on this page and found that \(-3-3i\) in polar form is \(3\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))\).

2. Identify the modulus and argument of the complex number. In this case, we have:

\(r=3\sqrt{2}, \ \theta=\frac{5\pi}{4}\)

3. Finally, we can substitute these values into the above formula and simplify:

\((3\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4})))^4\)

\(=(3\sqrt{2})^4(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))^4\)

\(=3^4(\sqrt{2})^4(cos(4(\frac{5\pi}{4}))+isin(4(\frac{5\pi}{4}))\)

\(=324(cos(5\pi)+isin(5\pi))\)

Notice that \(cos(5\pi)=-1\) and \(sin(5\pi)=0\). So, we can further simplify this to get \((-3-3i)^4=-324\).

- Last Updated: Apr 20, 2023 12:36 PM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717490
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