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Frank's questions at Yahoo! Answers regarding de Moivre's theorem

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MarkFL

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Feb 24, 2012
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Here are the questions:

Induction and Complex #'s. Calc help?


a) Prove, using mathematical induction, that for a positive integer n, (cos(x) + isinx)^n = cosnx +i sinnx where I^2 + -1

b)The complex number z is defined by z = cosx + isinx
I) show that 1/z = cos (-x) + isin(-x)
II) Deduce that z^n + z^-n = 2cosnx

c) Find the binomial expansion of (z + z^-1)^5
I) Hence show tat cos^5x = 1/16(acoas5x + bcos3x + ccosx) where a,b,c are positive integers to be found.

Thank you so much for you help!
I have posted a link there to this topic, so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Frank,

a) First, we check to see if $P_1$ (the case where $n=1$) is valid.

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^1= \cos(1 \cdot \theta)+i \cdot \sin(1 \cdot \theta)\)

\(\displaystyle \cos( \theta)+i \cdot \sin( \theta)= \cos( \theta)+i \cdot \sin( \theta)\)

$P_1$ is true. Next, our induction hypothesis $P_n$ is:

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^n= \cos(n \cdot \theta)+i \cdot \sin(n \cdot \theta)\)

Multiply both sides by \(\displaystyle \cos( \theta)+i \cdot \sin( \theta)\):

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^n \left( \cos( \theta)+i \cdot \sin( \theta) \right)= \left( \cos(n \cdot \theta)+i \cdot \sin(n \cdot \theta) \right) \left( \cos( \theta)+i \cdot \sin( \theta) \right)\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{n+1}= \cos(n \cdot \theta) \cos( \theta)+i \cdot \cos(n \cdot \theta) \sin( \theta)+i \cdot \cos( \theta) \sin(n \cdot \theta)+i^2 \cdot \sin( \theta) \sin(n \cdot \theta)\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{n+1}= \left( \cos(n \cdot \theta) \cos( \theta)- \sin(n \cdot \theta) \sin( \theta) \right)+i \left( \sin( \theta) \cos(n \cdot \theta)+ \cos( \theta) \sin(n \cdot \theta) \right)\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{n+1}= \cos \left(n \cdot \theta+ \theta \right)+i \cdot \sin \left(n \cdot \theta+ \theta \right)\)

\(\displaystyle \left( \cos(\theta)+i \cdot \sin( \theta) \right)^{n+1}= \cos \left((n+1) \cdot \theta \right)+i \cdot \sin \left((n+1) \cdot \theta \right)\)

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

b) For this question and for part c), we will find proving the following useful:

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-n}= \cos(n \cdot \theta)-i \cdot \sin(n \cdot \theta)\) where $n\le0\in\mathbb{Z}$

First, we check to see if $P_0$ (the case where $n=0$) is valid.

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^0= \cos(0 \cdot \theta)+i \cdot \sin(0 \cdot \theta)\)

\(\displaystyle 1= \cos(0)+i \cdot \sin(0)=1\)

$P_0$ is true. Next, our induction hypothesis $P_n$ is:

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-n}= \cos(n \cdot \theta)-i \cdot \sin(n \cdot \theta)\)

Multiply both sides by \(\displaystyle \left(\cos( \theta)+i \cdot \sin( \theta) \right)^{-1}\):

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-n} \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-1}= \frac{\cos(n \cdot \theta)-i \cdot \sin(n \cdot \theta)}{ \cos( \theta)+i \cdot \sin( \theta)}\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-(n+1)}= \frac{ \cos(n \cdot \theta)-i \cdot \sin(n \cdot \theta)}{ \cos( \theta)+i \cdot \sin( \theta)} \cdot \frac{ \cos( \theta)-i \cdot \sin( \theta)}{ \cos( \theta)-i \cdot \sin( \theta)}\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-(n+1)}= \frac{ \cos(n \cdot \theta) \cos(\theta)-i \cos(n \cdot \theta) \sin(\theta)-i \sin(n \cdot \theta) \cos(\theta)+i^2 \sin(n \cdot \theta) \sin(\theta)}{\cos^2(\theta)+\sin^2(\theta)}\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-(n+1)}= \left(\cos(n \cdot \theta) \cos(\theta)- \sin(n \cdot \theta) \sin(\theta) \right)-i \left(\cos(n \cdot \theta) \sin(\theta)+ \sin(n \cdot \theta) \cos(\theta) \right)\)

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-(n+1)}= \cos \left((n+1)\theta \right)-i \cdot \sin \left((n+1)\theta \right)\)

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

Observing that $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$, we may state:

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{-n}= \cos(-n \cdot \theta)+i \cdot \sin(-n \cdot \theta)\)

Thus, we have proved:

\(\displaystyle \left( \cos( \theta)+i \cdot \sin( \theta) \right)^{k}= \cos(k \cdot \theta)+i \cdot \sin(k \cdot \theta)\) where \(\displaystyle k\in\mathbb{Z}\)

i) Given that \(\displaystyle z=\cos(x)+i \cdot \sin(x)\) then we may use our theorem (actually de Moivre's theorem) to state:

\(\displaystyle \frac{1}{z}=z^{-1}=\left(\cos(x)+i \cdot \sin(x) \right)^{-1}=\cos(-x)+i \cdot \sin(-x)\)

ii) And so we may show that:

\(\displaystyle z^n+z^{-n}=\cos(nx)+i \cdot \sin(nx)+\cos(x)-i \cdot \sin(nx)=2 \cos(nx)\)

c) Using the binomial theorem, we find:

\(\displaystyle \left(z+z^{-1} \right)^5=\sum_{k=0}^5{5 \choose k}z^{5-k}z^{-k}=\sum_{k=0}^5{5 \choose k}z^{5-2k}\)

\(\displaystyle \left(z+z^{-1} \right)^5=z^5+5z^3+10z+10z^{-1}+5z^{-3}+z^{-5}\)

\(\displaystyle \left(z+z^{-1} \right)^5=\left(z^5+z^{-5} \right)+5\left(z^3+z^{-3} \right)+10\left(z+z^{-1} \right)\)

Using the result of b) ii) we may state:

\(\displaystyle \left(z+z^{-1} \right)^5=\left(2 \cos(5x) \right)+5\left(2 \cos(3x) \right)+10\left(2 \cos(x) \right)\)

\(\displaystyle \left(z+z^{-1} \right)^5=2 \cos(5x)+10 \cos(3x)+20 \cos(x)\)

i) Using the result of b) ii) we may state:

\(\displaystyle \left(z+z^{-1} \right)^5=\left(2\cos(x) \right)^5=32\cos^5(x)\)

And so using the previous result, we then find:

\(\displaystyle 32\cos^5(x)=2 \cos(5x)+10 \cos(3x)+20 \cos(x)\)

Hence:

\(\displaystyle \cos^5(x)=\frac{1}{16}\left( \cos(5x)+5 \cos(3x)+10 \cos(x) \right)\)

Thus, we have found:

\(\displaystyle a=1,\,b=5,\,c=10\)