Sketch $\displaystyle z^6 =1$ in an Argand Diagram
I have worked the values out to be:
$\displaystyle \frac\pi6,\frac\pi2,\frac{5\pi}{6},\frac{7\pi}{6}, \frac{3\pi}{2},\frac{11\pi}{6}$
I just do not know how to draw this.
Sketch $\displaystyle z^6 =1$ in an Argand Diagram
I have worked the values out to be:
$\displaystyle \frac\pi6,\frac\pi2,\frac{5\pi}{6},\frac{7\pi}{6}, \frac{3\pi}{2},\frac{11\pi}{6}$
I just do not know how to draw this.
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 $\displaystyle \frac {2 \pi}3$. this would be your $\displaystyle \theta$ in the diagram you see on the site i gave you. begin by measuring an angle of $\displaystyle \frac {2 \pi}3$ 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
indeed, but you have to look at what values your $\displaystyle \theta$ must take on.
note for instance, that $\displaystyle \theta = 2 \pi$ is one solution, since for this value, we have $\displaystyle z = 1$, which is clearly a solution to $\displaystyle z^6 = 1$
for $\displaystyle \theta = 6x$ you have the values, $\displaystyle 0,~2 \pi, ~4 \pi,~6 \pi, \cdots, 2k \pi, \cdots$
now take the values you need
This is the full question
How many complex numbers $\displaystyle z$ are there such that $\displaystyle 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 $\displaystyle \theta$ valuse or the x values?
Sorry to cause confusion
yes. an (non-constant) nth degree polynomial over the complex numbers has n roots (Fundamental Theorem of Algebra)
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)For this question do I sketch with $\displaystyle \theta$ valuse or the x values?
Sorry to cause confusion
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: $\displaystyle e^{i \theta} = \cos \theta + i \sin \theta$. working with $\displaystyle e^{i \theta}$ would have made the process easier to see.
anyway, you have
$\displaystyle z^6 = \cos x + i \sin x$
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 $\displaystyle 2 \pi$ is also a solution. thus, the solutions are of the form, $\displaystyle x = 0 + 2 k \pi$, or simply $\displaystyle x = 2 k \pi$, where $\displaystyle k$ is an integer.
so you have:
$\displaystyle z^6 = \cos 2k \pi + i \sin 2k \pi$
but now, to solve for $\displaystyle z$, simply divide the angles by 6. we have
$\displaystyle z = \cos \frac {2 k \pi}6 + i \sin \frac {2k \pi}6 = \cos \frac {k \pi}3 + i \sin \frac {k \pi}3$
now, let $\displaystyle k$ take on the values 0, 1, 2, 3, 4, and 5. (it will start to repeat after that. in general, for something like $\displaystyle z^n = \cos \theta + i \sin \theta$, you would proceed in this manner and let $\displaystyle k$ take on the values $\displaystyle 0,~1,~2, \cdots ,~n - 1$)
got it?
now finish up. i'm going to bed