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

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• November 27th 2012, 07:50 PM
ImConfused
Solve using de Moivre's theorem: z^3=i
Hi! So I have to find the solutions to the following equation using de Moivre's theorem: z^3=i

Here's my attempt at solving:

z^3=i
z^3=0+1i
r=(0^2+1^2)^1/2
r=1
arg=tan^-1(1/0)
arg=0

so z^3=1cis0

z1=1cis0
z2=1cis120
z3=1cis240

I feel like something is wrong with these solutions. (Worried)

• November 27th 2012, 08:16 PM
MarkFL
Re: Solve using de Moivre's theorem: z^3=i
I think I would write:

$z^3=i$

$z^3=0+i$

$z^3=\cos\left(\frac{\pi}{2}+2k\pi \right)+i\sin\left(\frac{\pi}{2}+2k\pi \right)$

$z=\left(\cos\left(\frac{\pi}{2}(4k+1) \right)+i\sin\left(\frac{\pi}{2}(4k+1) \right) \right)^{\frac{1}{3}}$

$z=\cos\left(\frac{\pi}{6}(4k+1) \right)+i\sin\left(\frac{\pi}{6}(4k+1) \right)$

where $k\in\{0,1,2\}$ hence:

$z_1=\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_2=-\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_3=-i$
• November 27th 2012, 08:47 PM
Deveno
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by ImConfused
Hi! So I have to find the solutions to the following equation using de Moivre's theorem: z^3=i

Here's my attempt at solving:

z^3=i
z^3=0+1i
r=(0^2+1^2)^1/2
r=1
arg=tan^-1(1/0)
arg=0

so z^3=1cis0

z1=1cis0
z2=1cis120
z3=1cis240

I feel like something is wrong with these solutions. (Worried)

arg(i) = π/2. we have to make an exception for i and -i because tan-1(1/0) and tan-1(-1/0) are undefined (the tangent of π/2 is infinite, and the tangent of -π/2 is negative infinite).

but just looking at the complex plane, it is obvious the y-axis (the imaginary axis) points "straight-up" (90 degrees counterclockwise to the x-axis).

so i = cis(π/2), NOT 0. this should give you more "reasonable" answers for z1, z2 and z3, namely:

z1 = cis(π/6) <--its good to get used to thinking in terms of pi. you can write cis(30), if you prefer.
z2 = cis(7π/6) (or cis(150) which is "a full circle plus 90 degees (450 degrees) divided by 3")
z3 = cis(3π/2) = cis(-π/2) (or cis(270) = cis(-90), depending on what your "principal range of angles" is. note that 270 is one third of "two full circles plus 90 degrees": 720+90 = 810, and 810/3 = 270).

if you insist on evaluating cos(π/6) + i sin(π/6) (and so on with the other two cube roots), you get the same thing as MarkFL2.
• November 27th 2012, 08:58 PM
ImConfused
Re: Solve using de Moivre's theorem: z^3=i
Thanks MarkFL2. I can follow what you're saying but I didn't learn complex numbers with some of the techniques you used. Thanks again though :)

Deveno, I am kind of confused by the use of n. Does n represent the power to which the complex number is raised? And is arg(i)=n/2 a general rule that I should know?
• November 27th 2012, 09:03 PM
MarkFL
Re: Solve using de Moivre's theorem: z^3=i
Deveno has used the pi character "π", not the letter "n".
• November 27th 2012, 09:09 PM
Deveno
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by ImConfused
Thanks MarkFL2. I can follow what you're saying but I didn't learn complex numbers with some of the techniques you used. Thanks again though :)

Deveno, I am kind of confused by the use of n. Does n represent the power to which the complex number is raised? And is arg(i)=n/2 a general rule that I should know?

no it is supposed to be the symbol for "pi", which does not look very good in the default sans serif font used on these forums. i have changed the font to make it more recognizable.
• November 27th 2012, 09:26 PM
ImConfused
Re: Solve using de Moivre's theorem: z^3=i
Oh! That makes much more sense!

Thanks soo much for your help :)
• January 12th 2013, 01:47 PM
HallsofIvy
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by ImConfused
Hi! So I have to find the solutions to the following equation using de Moivre's theorem: z^3=i

Here's my attempt at solving:

z^3=i
z^3=0+1i
r=(0^2+1^2)^1/2
r=1
arg=tan^-1(1/0)
arg=0

This is wrong. tan(0) is 0, not "1/0" which does not actually exist, though you can think of it as " $+\infty$". What is true is that as $\theta$ goes to $\pi/2$, $\tan(\theta)$ increases without bound ("goes to infinity"). In any case, you should recognise that, marking "i" on the complex plane, the angle is $\pi/2$, not 0.

Quote:

so z^3=1cis0
No, as pointed out above. $z^3= cis(\pi/2)$. You seem to be relying on blindly applying (or trying to apply) formulas without thinking about what these things mean. Of course r= |i|= 1, you shouldn't have to do " $\sqrt{0^2+ 1^2}= 1$" to see that.

Quote:

z1=1cis0
z2=1cis120
z3=1cis240

I feel like something is wrong with these solutions. (Worried)
Those are the cube roots of 1, not i.
• January 19th 2015, 03:16 AM
Martinun
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by MarkFL
I think I would write:

$z^3=i$

$z^3=0+i$

$z^3=\cos\left(\frac{\pi}{2}+2k\pi \right)+i\sin\left(\frac{\pi}{2}+2k\pi \right)$

$z=\left(\cos\left(\frac{\pi}{2}(4k+1) \right)+i\sin\left(\frac{\pi}{2}(4k+1) \right) \right)^{\frac{1}{3}}$

$z=\cos\left(\frac{\pi}{6}(4k+1) \right)+i\sin\left(\frac{\pi}{6}(4k+1) \right)$

where $k\in\{0,1,2\}$ hence:

$z_1=\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_2=-\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_3=-i$

Can you please tell me how you get 4k+1 from 2kpi ?
• January 19th 2015, 03:34 AM
Prove It
Re: Solve using de Moivre's theorem: z^3=i
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*}
• January 19th 2015, 04:36 AM
Martinun
Re: Solve using de Moivre's theorem: z^3=i
Quote:

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*}

Thank you can you try solve for example z^4 = -16i or z^5 = 2+2i this way ?
• January 19th 2015, 04:46 AM
Prove It
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by Martinun
Thank you can you try solve for example z^4 = -16i or z^5 = 2+2i this way ?

Absolutely, just put each of those complex numbers in their most general exponential form, and then you should be able to apply your index laws to solve the equation.

\displaystyle \begin{align*} z^4 &= -16i \\ z^4 &= 16\mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ 16\mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{4}} \\ z &= 16^{\frac{1}{4}}\left[ \mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{4}} \\ z &= 2\mathrm{e}^{ \left( -\frac{\pi}{8} + \frac{\pi}{2}n \right) \mathrm{i} } \end{align*}

See if you can get all the Cartesian forms of the roots from the first cycle. See if you can do the second question...
• January 19th 2015, 04:48 AM
SlipEternal
Re: Solve using de Moivre's theorem: z^3=i
\begin{align*}z^4 & = -16i \\ z^4 & = 16e^{ \left( \tfrac{\pi}{2} + 2\pi n \right) i } \\ z & = \left[ 16 e^{ \left( \tfrac{\pi}{2} + 2\pi n\right) i }\right]^{\tfrac{1}{4}} \\ z & = 2 e^{ \left( \tfrac{\pi+4\pi n}{8} \right) i }\end{align*}

Can you solve it from here?

Edit: Prove It beat me to it.
• January 19th 2015, 04:56 AM
Prove It
Re: Solve using de Moivre's theorem: z^3=i
I also realise I made a mistake in my first post. All those exponentials need to have "i" in its power.
• January 19th 2015, 05:32 AM
skeeter
Re: Solve using de Moivre's theorem: z^3=i
Quote:

Originally Posted by Martinun
Can you please tell me how you get 4k+1 from 2kpi ?

To answer your original question, note that $\frac{\pi}{2}$ was factored out of each term ...

$\frac{\pi}{2} + 2k \pi = \frac{\pi}{2} \cdot 1 + \frac{\pi}{2} \cdot 4k = \frac{\pi}{2} \left(1 + 4k \right)$
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