January 11th, 2018, 06:28 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 80 Thanks: 0  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 \displaystyle z = 2^{\frac{1}{6}} e^{i\frac{\pi}{12}} $ $\displaystyle \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. 
January 11th, 2018, 07:14 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
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). Last edited by v8archie; January 11th, 2018 at 07:17 AM. 
January 11th, 2018, 09:05 AM  #3 
Senior Member Joined: Aug 2012 Posts: 1,922 Thanks: 534 
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.

January 12th, 2018, 04:28 AM  #4 
Newbie Joined: Jan 2018 From: Tunis, Tunisia Posts: 10 Thanks: 4 Math Focus: Analysis 
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 $\displaystyle j^2=\overline j=\dfrac{1i\sqrt3}2=e^{i\frac{2\pi}3}$. This result can be explained by : $\displaystyle z^3=w=z_0^3\iff\left(\dfrac{z}{z_0}\right)^3=1\iff \dfrac{z}{z_0}\in\{1,j,j^2\}$. 
January 12th, 2018, 05:35 AM  #5 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography 
Salam ! We can use Euler's formula, any nonzero 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 Then we can get :\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} $ $\displaystyle r=\sqrt[6]{2}\quad\theta\in\{\dfrac\pi{12};\dfrac{3\pi}4; \dfrac{17\pi}{12}\}.$ Last edited by skipjack; January 12th, 2018 at 12:06 PM. 
January 12th, 2018, 11:38 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619  
January 12th, 2018, 11:57 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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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