Thread: Solve using de Moivre's theorem: z^3=i

1. Re: Solve using de Moivre's theorem: z^3=i

Originally Posted by Prove It
First of all, when dealing with solving exponential equations, the EXPONENTIAL form of the complex number is ALWAYS easiest to use...

\displaystyle \begin{align*} z^3 &= \mathrm{i} \\ z^3 &= \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{3}} \\ z &= \mathrm{ e }^{ \frac{\pi}{6} + \frac{2\pi\,n}{3} } \end{align*}

So in the first cycle, the possible solutions are \displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}}, \mathrm{e}^{\frac{5\pi}{6}} , \mathrm{e}^{\frac{3\pi}{2}} \end{align*}, which in Cartesian form are:

\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*}

I am really bad in this. Sorry guys. I cant really solve it. Can you try solve whole example and like for real bad student ?
For example how did you get from i e on something ?
If someone have time would be nice if he can solve this examples but every step what you do..
I have like 4-5 examples and cant really solve it. and i have exam in 2 days
Best would be way like" Prove it " solved it. We are learning it like that.
z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
or is there some manual where they describe everything.
Anyway thats for help what you allready solved.

2. Re: Solve using de Moivre's theorem: z^3=i

If you can put your equation in "cis" form, then you can put the equation in exponential form. In general \displaystyle \begin{align*} r\,\mathrm{e}^{\mathrm{i}\,\theta} \equiv r \left[ \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \right] \end{align*}

3. Re: Solve using de Moivre's theorem: z^3=i

Originally Posted by Martinun
For example how did you get from i e on something ?
If someone have time would be nice if he can solve this examples but every step what you do..
I have like 4-5 examples and cant really solve it. and i have exam in 2 days
Best would be way like" Prove it " solved it. We are learning it like that.
z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
You are a victim of poor mathematical education practice. There is a good trend to use a more geometric approach to teaching complex analysis.

First you need to understand that any complex has n nth roots. Those n roots are located on a circle centered at the origin of radius the nth root of absolute value of the number. Those n roots divide the circle into n equal arcs.

Let's stop a moment for an example. There are six sixth roots of $3-4\bf{i}$. They are located on a circle centered at $(0,0)$ and radius $\sqrt[6]5$. On that circle the roots are separated by arcs of $\dfrac{2\pi}{6}$, where $2\pi$ is the whole circle and six equal parts.

Now give some notation to simplify the way we can list the roots.
Let $\exp(i\theta)=\cos(\theta)+i\sin(\theta)$.
Then for a complex number $z=x+yi$ then $\exp(z)=e^x\left(\exp(iy)\right)=\sqrt{x^2+y^2}\bf {cis}(\theta)$.
Finally we give the notation for the argument of the complex number:
Suppose that $x\cdot y\ne 0$ then $\theta= Arg(x + yi) = \left\{ {\begin{array}{{lr}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.$

Here is a example of the process: $z^5=2-2i$. $\bf{Arg}(2-2i})=\theta=\arctan\left(\frac{-2}{2}\right)$
This is the principle root, $\rho=\sqrt[5]{|2-2i|}\exp\left(\dfrac{i\theta}{5}\right)$, so that $\rho^5=2-2i$

List all the roots. Let $\zeta=\exp\left(\frac{2\pi}{5}~\bf{i}\right)$
the five roots are: $\rho\cdot\zeta^k,~k=0,1,2,3,4$

4. Re: Solve using de Moivre's theorem: z^3=i

Originally Posted by Martinun
Can you please tell me how you get 4k+1 from 2kpi ?
MarkFL didn't get 4k+ 1 from $2k\pi$.

What he had was $\frac{\pi}{2}+ 2k\pi$. Getting the "common denominator", that is $\frac{\pi}{2}+ \frac{4k\pi}{2}$ $= \frac{\pi+ 4k\pi}{2}= \frac{\pi}{2}(4k+ 1)$.

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using de moivre's theorem to show that z^3 1 =0

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