Question

Use DeMoivre's Theorem to solve (3+sqrt(3i))^4

Answer 1

your job is to write
\[3+\sqrt{3}i\] in the form
\[r\cos(\theta)+i\sin(\theta)\] and then it will be ral easy

Answer 2

First, figure out whether you can rewrite \(3+i\sqrt3\) as an expression of the form \(r(\cos\theta+i\sin\theta)\). DeMoivre's theorem tells us \([r(\cos\theta+i\sin\theta)]^n=r^n(\cos n\theta+i\sin n\theta)\) for integer \(n\).

Answer 3

make that
\[r\left(\cos(\theta)+i\sin(\theta)\right)\]

Answer 4

Note that:
\[
(3+i\sqrt{3})=2\sqrt{3}e^{i\frac{\pi}{6}}
\]Taking this to the 4th power:
\[
(3+i\sqrt{3})^4=(2\sqrt{3})^4e^{i4\frac{\pi}{6}}=144e^{i\frac{2\pi}{3}}=144(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3})
\]So:
\[
144(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3})=144(-\frac{1}{2}+\frac{\sqrt{3}}{2})=-72+72\sqrt{3}
\]

Answer 5

@LolWolf slow down mate...

Answer 6

you find \(r\) via \[r=\sqrt{a^2+b^2}\] and \(\theta\) via\[\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\] and \[\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}\]

Answer 7

Sorry; been typing that up for some time.

Answer 8

you guys are amazing!

Answer 9

@audreynicole47 @LolWolff observed that we may rewrite our expression as follows: $$3+i\sqrt3=2\left(\frac32+i\frac{\sqrt3}2\right)=\frac2{\sqrt3}\left(\frac{\sqrt3}2+i\frac12\right)=\frac2{\sqrt3}\left(\cos\frac\pi6+i\sin\frac\pi6\right)$$
You may verify this is correct; one way to determine this in a more straight-forward fashion is provided by @satellite73. The last step was just identifying what \(\theta\) will give you a cosine of \(\frac{\sqrt3}2\) and a sine of \(\frac12\); \(\frac\pi6\) is that angle, which is clear after a quick glance at the unit circle.

Answer 10

Now that we have rewritten our expression in the above form (typically known as *polar* form) we may apply DeMoivre's theorem:$$\left[\frac2{\sqrt3}\left(\cos\frac\pi6+i\sin\frac\pi6\right)\right]^4=\left(\frac2{\sqrt3}\right)^4\left(\cos\frac{4\pi}6+i\sin\frac{4\pi}6\right)$$... which we may reduce by recognizing \(\left(\dfrac2{\sqrt3}\right)^4=\dfrac{2^4}{\sqrt3^4}=\dfrac{16}9\)

Answer 11

Further, we see that \(\dfrac{4\pi}6=\dfrac{2\pi}3\) which may not reduced further. Thus our final result is merely:$$\frac{16}9\left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right)$$

Answer 12

oldrin.bataku....Thanks. Would it be a different tactic on how to solve an equation, again using DeMoivre's Theorem, but for (1-i)^-8? Since the power is negative?

Answer 13

@audreynicole47 nope! DeMoivre's theorem holds for negative integer \(n\) (and, as it turns out, non-integer rational \(n\) as well, though it's multivalued -- you'll see this when evaluating roots).

Answer 14

@oldrin.bataku Oy, be careful on your radius, there.

And also DeMoivre's theorem is only for integers, Euler's is for all complex numbers (which is essentially an 'extension' although it's quite different).

Answer 15

@audreynicole47 I'd like to show you something very neat that LolWolff alluded to. As it turns out, there is a very interesting relationship between the exponential function \(e^x\) (also known as \(\exp(x)\)) and the familiar trigonometric functions \(\cos\theta\) and \(\sin\theta\). Euler discovered the following formula:$$e^{i\theta}=\cos\theta+i\sin\theta$$What this means is that \(e^{i\theta}\) plots out the unit circle in the complex plane:
Because of this, it follows that once you rewrite our complex number in polar form, we observe $$2\sqrt3\left(\cos\frac\pi6+i\sin\frac\pi6\right)=2\sqrt3e^{i\frac\pi6}$$In this form, observe what happens when we take to the fourth power:$$\left(2\sqrt3e^{i\dfrac\pi6}\right)^4=144e^{i\dfrac{4\pi}6}=144\left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right)$$We see that Euler's formula actually generalizes DeMoivre's theorem for all real \(n\):$$\left[r\left(\cos\theta+i\sin\theta\right)\right]^n=\left(re^{i\theta}\right)^n=r^ne^{in\theta}=r^n\left(\cos n\theta+i\sin n\theta\right)$$