I think I would write:
where hence:
Hi! So I have to find the solutions to the following equation using de Moivre's theorem: z^3=i
Here's my attempt at solving:
z^3=i
z^3=0+1i
r=(0^2+1^2)^1/2
r=1
arg=tan^-1(1/0)
arg=0
so z^3=1cis0
z1=1cis0
z2=1cis120
z3=1cis240
I feel like something is wrong with these solutions.
HELP please!
arg(i) = π/2. we have to make an exception for i and -i because tan^{-1}(1/0) and tan^{-1}(-1/0) are undefined (the tangent of π/2 is infinite, and the tangent of -π/2 is negative infinite).
but just looking at the complex plane, it is obvious the y-axis (the imaginary axis) points "straight-up" (90 degrees counterclockwise to the x-axis).
so i = cis(π/2), NOT 0. this should give you more "reasonable" answers for z_{1}, z_{2} and z_{3}, namely:
z_{1} = cis(π/6) <--its good to get used to thinking in terms of pi. you can write cis(30), if you prefer.
z_{2} = cis(7π/6) (or cis(150) which is "a full circle plus 90 degees (450 degrees) divided by 3")
z_{3} = cis(3π/2) = cis(-π/2) (or cis(270) = cis(-90), depending on what your "principal range of angles" is. note that 270 is one third of "two full circles plus 90 degrees": 720+90 = 810, and 810/3 = 270).
if you insist on evaluating cos(π/6) + i sin(π/6) (and so on with the other two cube roots), you get the same thing as MarkFL2.
Thanks MarkFL2. I can follow what you're saying but I didn't learn complex numbers with some of the techniques you used. Thanks again though
Deveno, I am kind of confused by the use of n. Does n represent the power to which the complex number is raised? And is arg(i)=n/2 a general rule that I should know?
This is wrong. tan(0) is 0, not "1/0" which does not actually exist, though you can think of it as " ". What is true is that as goes to , increases without bound ("goes to infinity"). In any case, you should recognise that, marking "i" on the complex plane, the angle is , not 0.
No, as pointed out above. . You seem to be relying on blindly applying (or trying to apply) formulas without thinking about what these things mean. Of course r= |i|= 1, you shouldn't have to do " " to see that.so z^3=1cis0
z1=1cis0
z2=1cis120
z3=1cis240
I feel like something is wrong with these solutions.
Those are the cube roots of 1, not i.
HELP please!
First of all, when dealing with solving exponential equations, the EXPONENTIAL form of the complex number is ALWAYS easiest to use...
$\displaystyle \begin{align*} z^3 &= \mathrm{i} \\ z^3 &= \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{3}} \\ z &= \mathrm{ e }^{ \frac{\pi}{6} + \frac{2\pi\,n}{3} } \end{align*}$
So in the first cycle, the possible solutions are $\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}}, \mathrm{e}^{\frac{5\pi}{6}} , \mathrm{e}^{\frac{3\pi}{2}} \end{align*}$, which in Cartesian form are:
$\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*}$
Absolutely, just put each of those complex numbers in their most general exponential form, and then you should be able to apply your index laws to solve the equation.
$\displaystyle \begin{align*} z^4 &= -16i \\ z^4 &= 16\mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ 16\mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{4}} \\ z &= 16^{\frac{1}{4}}\left[ \mathrm{e}^{\left( -\frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{4}} \\ z &= 2\mathrm{e}^{ \left( -\frac{\pi}{8} + \frac{\pi}{2}n \right) \mathrm{i} } \end{align*}$
See if you can get all the Cartesian forms of the roots from the first cycle. See if you can do the second question...