# Plotting numbers in the complex plane?

• Mar 30th 2011, 04:02 AM
cottontails
Plotting numbers in the complex plane?
(I'm unsure as to whether this question can be regarded as calculus, however, I am currently studying Differential Calculus at university as a first year subject so, I just made the assumption there - this question came from my course's textbook.)

"Find all solutions of z^4 = 8 + 8√3i, and plot them in the complex plane."

First of all, I'm unsure with how you would be able to find the solutions for the equation. I understand that there are 4 roots to the equation, one is for linear equations in z and two for quadratics in z^2 and so on. I know that you need to express z^4 = 8 + 8√3i into the form r(cost + isint) and then use de Moivre's theorem from there. So, from doing that - how would you be able to get the points that need to be plotted? Although, I'm still confused as to how you would work out the entire question. This question was given as an example from my textbook however, I wasn't able to follow on with it quite clearly. What I am most particularly confused about is how you would plot the points. Apparently you can find one solution given that the other solutions of z^n = k form a regular polygon around the origin (which confuses me even more...) So, is anyone able to provide a helpful step-by-step procedure with how to get the answer?

Any help would be greatly appreciated. Thank you.
• Mar 30th 2011, 04:20 AM
FernandoRevilla
$\displaystyle 8+8\sqrt{3}i=16[\cos (\pi/3)+i\sin(\pi/3)]$.

Now, apply:

$\displaystyle \sqrt[4]{8+8\sqrt{3}i}=2\left[\cos\left(\dfrac{\pi}{12}+\dfrac{2k\pi}{4}\left)+i \sin\left(\dfrac{\pi}{12}+\dfrac{2k\pi}{4}\right)\ right],\quad (k=0,1,2,3)$
• Mar 30th 2011, 04:26 AM
Plato
Quote:

Originally Posted by cottontails
"Find all solutions of z^4 = 8 + 8√3i, and plot them in the complex plane."

First $\displaystyle 8+8\sqrt{3}=16(\text{cis}\left(\frac{\pi}{3}\right ))$
One fourth root of that is clearly $\displaystyle 2(\text{cis}\left(\frac{\pi}{12}\right))$.
Now the other three roots are equally distributed about the circle at angles of $\displaystyle \dfrac{2\pi}{4}$ apart.
So another root is $\displaystyle 2(\text{cis}\left(\frac{7\pi}{12}\right))$.
Can you find the other two?
• Mar 30th 2011, 04:49 AM
cottontails
I was able to figure that when k = 0, 2(cis(pi/12)).
However my answers for the other values of k:
k = 1, 2(cis(7pi/12))
k = 2, 2(cis(13pi/12))
k = 3, 2(cis(19pi/12))
So, none of my other roots matched up as being 2(cis(5pi/12)). So, I'm pretty sure I'm doing something wrong there. What I did was basically sub-in the values of k from FernandoRevilla's answer: 2(cos(pi/12 + 2kpi/4) + isin(pi/12 + 2kpi/4)) Another similar example from my textbook did the same thing and its answers were correct. So, I'm unsure with what I did wrong. Or, could it be something that k = 1 is 2(cis(5pi/12))? Although then again, I don't know how that would work out with the other values of k.
• Mar 30th 2011, 05:57 AM
Plato
That typo has been corrected.
• Mar 30th 2011, 06:35 AM
cottontails
Apparently the values plotted on a complex plane lie at the corners of a square. However, I really don't have a clue with how can I possibly plot these values on the complex plane. Would I have to change it back into cartesian form to be able to plot all the values on the complex plane then?
• Mar 30th 2011, 07:04 AM
Plato
Quote:

Originally Posted by cottontails
Apparently the values plotted on a complex plane lie at the corners of a square. However, I really don't have a clue with how can I possibly plot these values on the complex plane. Would I have to change it back into cartesian form to be able to plot all the values on the complex plane then?

As I said above, the numbers are on a circle.
In the first graphic the four fourth roots are plotted.
In the second graphic the six sixth roots are plotted.
• Apr 5th 2011, 08:21 PM
cottontails
Thank you! That solves my question. :)