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Math Help - Solve using de Moivre's theorem: z^3=i

  1. #1
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    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.

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

    Quote Originally Posted by ImConfused View Post
    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.

    HELP please!
    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.
    Last edited by Deveno; November 27th 2012 at 09:07 PM.
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    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?
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    Re: Solve using de Moivre's theorem: z^3=i

    Deveno has used the pi character "π", not the letter "n".
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by ImConfused View Post
    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.
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    Re: Solve using de Moivre's theorem: z^3=i

    Oh! That makes much more sense!

    Thanks soo much for your help
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by ImConfused View Post
    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.


    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.

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

    I feel like something is wrong with these solutions.
    Those are the cube roots of 1, not i.
    HELP please!
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by MarkFL View Post
    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 ?
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    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:

    $\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*}$
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Prove It View Post
    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 ?
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Martinun View Post
    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...
    Last edited by Prove It; January 19th 2015 at 04:57 AM.
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    Re: Solve using de Moivre's theorem: z^3=i

    For your first example:
    \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.
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    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.
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    Re: Solve using de Moivre's theorem: z^3=i

    Quote Originally Posted by Martinun View Post
    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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