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January 11th, 2018, 06:28 AM   #1
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Question Solve for z^3 = 1 + i

Tried using Symbolabs Algebra calculator. Didn't work out well since it said there are currently no solutions to this equation but I know there is.

My answer isn't correct by the way. One of the answers were
\displaystyle z = 2^{\frac{1}{6}} e^{i\frac{\pi}{12}}

\displaystyle z^{3} = 1 + i\\
de \ Moivre's \ formula\\
\displaystyle re^{i\theta} = r(\cos(\theta) + i\sin(\theta)\\
\displaystyle |r| = \sqrt{a^{2} + b^{2}}\\
\displaystyle |r| = \sqrt{2}\\
\displaystyle 1\cos(\theta) = 1 \Rightarrow \theta = \pm 0 + 2\pi n\\
\displaystyle 1\sin(\theta) = 1 \Rightarrow \theta = \frac{\pi}{2} + 2\pi n \\
\displaystyle re^{i\theta} = \sqrt{2}e^{i\frac{\pi}{2}}\\
\displaystyle z^3 = \sqrt{2}e^{i\frac{\pi}{2}}\\
\displaystyle z = \sqrt{2}^{\frac{1}{3}} e^{i\frac{\pi}{6}}

Last edited by DecoratorFawn82; January 11th, 2018 at 06:35 AM.
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January 11th, 2018, 07:14 AM   #2
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It's not $1\cos\theta = 1$, it should be $\sqrt{2}\cos\theta = 1$.
Similarly, it should be $\sqrt2\sin\theta = 1$.

It's quite easy to verify your value of theta by thinking about where $1 + \mathrm i$ is on the complex plane. A moment's thought should convince you that $\theta$ cannot be $\frac\pi2$ (on the imaginary axis).
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Last edited by v8archie; January 11th, 2018 at 07:17 AM.
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January 11th, 2018, 09:05 AM   #3
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Always start these problems by drawing a picture. $1 + i$ corresponds to the point $(1,1)$ on the plane. That makes an angle of $\frac{\pi}{4}$ with the positive $x$-axis. One third of that is $\frac{\pi}{12}$. Then you just have to account for the length and the other two roots, which are powers of the one you just found.
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January 12th, 2018, 04:28 AM   #4
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Good morning !

In general, if we wont to solve the equation : $\displaystyle z^3=w$ and if we know a solution $\displaystyle z_0$, then the other two solutions are :
$\displaystyle z_1=jz$ and $\displaystyle z_2=j^2z$,
where $\displaystyle j=\dfrac{-1+i\sqrt3}2=e^{i\frac{2\pi}3}$ is the imaginary cube root of the unity.

We have $\displaystyle j^2=\overline j=\dfrac{-1-i\sqrt3}2=e^{-i\frac{2\pi}3}$.

This result can be explained by :
$\displaystyle z^3=w=z_0^3\iff\left(\dfrac{z}{z_0}\right)^3=1\iff \dfrac{z}{z_0}\in\{1,j,j^2\}$.
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January 12th, 2018, 05:35 AM   #5
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Salam !

We can use Euler's formula, any non-zero complex number $\displaystyle z$ can be written as $\displaystyle z=re^{i\theta}$, where $\displaystyle r=|z|$ and $\displaystyle \theta=\arg z$. (Even $\displaystyle 0$ can be written in that way if we chose $\displaystyle r=0$ and in this case $\displaystyle \theta$ can be any real number).

Let $\displaystyle z=re^{i\theta}$ a solution of the equation $\displaystyle z^3=1+i$. We can write $\displaystyle 1+i=\sqrt2e^{i\frac\pi4}$. So we get :
$\displaystyle z^3=1+i\iff r^3e^{3i\theta}=\sqrt2e^{i\frac\pi4}\iff
\begin{cases}r^3=\sqrt2\\3\theta=\dfrac \pi4+2k\pi\end{cases}\iff
\begin{cases}r=2^{\frac16}\\\theta=\dfrac \pi{12}+
\dfrac{2k\pi}3\quad k\in\{0,1,2\}\end{cases}
Then we can get :
$\displaystyle r=\sqrt[6]{2}\quad\theta\in\{\dfrac\pi{12};\dfrac{3\pi}4;
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Last edited by skipjack; January 12th, 2018 at 12:06 PM.
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January 12th, 2018, 11:38 AM   #6
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Originally Posted by Maschke View Post
. . . and the other two roots, which are powers of the one you just found.
That's not quite right. All the roots have the same modulus.
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January 12th, 2018, 11:57 AM   #7
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One of the roots is $\displaystyle \frac{1 + √3}{2{\large∛}2} + \frac{√3 - 1}{2{\large∛}2}i$.

Another is $\displaystyle -\frac{1}{{\large∛}2} + \frac{1}{{\large∛}2}i$.
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