# [SOLVED] Sketch z^6 = 1 on an Argand diagram

• January 18th 2009, 08:54 PM
ronaldo_07
[SOLVED] Sketch z^6 = 1 on an Argand diagram
Sketch $z^6 =1$ in an Argand Diagram

I have worked the values out to be:

$\frac\pi6,\frac\pi2,\frac{5\pi}{6},\frac{7\pi}{6}, \frac{3\pi}{2},\frac{11\pi}{6}$

I just do not know how to draw this.
• January 18th 2009, 09:17 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
Sketch z6 =1 in an Argand Diagram

I have worked the values out to be:

$\frac\pi6,\frac\pi2,\frac{5\pi}{6},\frac{7\pi}{6}, \frac{3\pi}{2},\frac{11\pi}{6}$

I just do not know how to draw this.

i suppose you mean $z^6 = 1$ ??

in that case, your values are wrong.

when you find the correct ones (can you?), see if you can continue after looking at this
• January 18th 2009, 09:22 PM
ronaldo_07
Quote:

Originally Posted by Jhevon
i suppose you mean $z^6 = 1$ ??

in that case, your values are wrong.

when you find the correct ones (can you?), see if you can continue after looking at this

yes thats the formulae

I have checked again is the first value $\frac{\pi}{12}$

And I am unsure How I use my values to create an argand diagram
• January 18th 2009, 09:36 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
yes thats the formulae

I have checked again is the first value

And I am unsure How I use my values to create an argand diagram

first draw (faintly) a circle of radius 1. you are going to draw the points that correspond to each complex number on that circle. the x and y coordinates correspond to the real and imaginary parts of the complex number respectively. the angles you find (called "arguments") tell you how to locate the complex number based on its angle from the positive x-axis.

example, one of the answers is $\frac {2 \pi}3$. this would be your $\theta$ in the diagram you see on the site i gave you. begin by measuring an angle of $\frac {2 \pi}3$ anti-clockwise from the positive x-axis, indicating in with a line going from the origin as you see in the diagram. then, wherever it touches the circle, that's the complex number. plot a point there, and you're done
• January 18th 2009, 09:38 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
I have checked again is the first value $\frac{\pi}{12}$

...how exactly are you coming up with these values? notice that you are dividing the circle into 6 equal slices, the first angle being zero (corresponding to the solution z = 1)
• January 18th 2009, 09:40 PM
ronaldo_07
$z=cos(x)+isin(x)$

$z^6$ = $cos(6x)+isin(6x)$

because I need $6x=\theta$ I divide by 6

was I wrong to do this?
• January 18th 2009, 10:05 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
$z=cos(x)+isin(x)$

$z^6$ = $cos(6x)+isin(6x)$

because i need $6x=\theta$ I divide by 6

indeed, but you have to look at what values your $\theta$ must take on.

note for instance, that $\theta = 2 \pi$ is one solution, since for this value, we have $z = 1$, which is clearly a solution to $z^6 = 1$

for $\theta = 6x$ you have the values, $0,~2 \pi, ~4 \pi,~6 \pi, \cdots, 2k \pi, \cdots$

now take the values you need
• January 18th 2009, 10:16 PM
ronaldo_07
This is the full question

How many complex numbers $z$ are there such that $z^6 =1?$Calculate them all and sketch them in the Argand diagram.

I think there are 6 complex numbers. is this correct?

For this question do I sketch with $\theta$ valuse or the x values?

Sorry to cause confusion
• January 18th 2009, 10:35 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
This is the full question

How many complex numbers $z$ are there such that $z^6 =1?$Calculate them all and sketch them in the Argand diagram.

I think there are 6 complex numbers. is this correct?

yes. an (non-constant) nth degree polynomial over the complex numbers has n roots (Fundamental Theorem of Algebra)

Quote:

For this question do I sketch with $\theta$ valuse or the x values?

Sorry to cause confusion
i figured the question was something like that. you should have posted the question in its entirety to begin with though. (i thought you would have learned your lesson from here -- post #11 made a particularly suggestive comment)

i will continue using the form you have shown me, but i believe another form would be easier to work with, that is, Euler's equation: $e^{i \theta} = \cos \theta + i \sin \theta$. working with $e^{i \theta}$ would have made the process easier to see.

anyway, you have

$z^6 = \cos x + i \sin x$

you want this to equal 1. you should note two things. first, note one solution, the most basic and obvious one is the one you should think of (here, that would be x = 0). the second thing you should note, is that if an angle is a solution, then that angle plus any multiple of $2 \pi$ is also a solution. thus, the solutions are of the form, $x = 0 + 2 k \pi$, or simply $x = 2 k \pi$, where $k$ is an integer.

so you have:

$z^6 = \cos 2k \pi + i \sin 2k \pi$

but now, to solve for $z$, simply divide the angles by 6. we have

$z = \cos \frac {2 k \pi}6 + i \sin \frac {2k \pi}6 = \cos \frac {k \pi}3 + i \sin \frac {k \pi}3$

now, let $k$ take on the values 0, 1, 2, 3, 4, and 5. (it will start to repeat after that. in general, for something like $z^n = \cos \theta + i \sin \theta$, you would proceed in this manner and let $k$ take on the values $0,~1,~2, \cdots ,~n - 1$)

got it?

now finish up. i'm going to bed (Yawn)
• January 18th 2009, 10:46 PM
ronaldo_07
Quote:

Originally Posted by Jhevon
i figured the question was something like that. you should have posted the question in its entirety to begin with though. (i thought you would have learned your lesson from here -- post #11 made a particularly suggestive comment)

Your right I only posted the part of the question that I had not done (the sketch) so that I do not cause confusion. But it appears that had backfired (Worried)