Solve using de Moivre's theorem: z^3=i

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November 27th 2012, 07:50 PM
ImConfused

Solve using de Moivre's theorem: z^3=i

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. (Worried)

HELP please!
November 27th 2012, 08:16 PM
MarkFL

Re: Solve using de Moivre's theorem: z^3=i

I think I would write:

$z^3=i$

$z^3=0+i$

$z^3=\cos\left(\frac{\pi}{2}+2k\pi \right)+i\sin\left(\frac{\pi}{2}+2k\pi \right)$

$z=\left(\cos\left(\frac{\pi}{2}(4k+1) \right)+i\sin\left(\frac{\pi}{2}(4k+1) \right) \right)^{\frac{1}{3}}$

$z=\cos\left(\frac{\pi}{6}(4k+1) \right)+i\sin\left(\frac{\pi}{6}(4k+1) \right)$

where $k\in\{0,1,2\}$ hence:

$z_1=\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_2=-\frac{\sqrt{3}}{2}+\frac{1}{2}i$

$z_3=-i$
November 27th 2012, 08:47 PM
Deveno

Re: Solve using de Moivre's theorem: z^3=i

Quote:

Originally Posted by ImConfused

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. (Worried)

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 z₁, z₂ and z₃, namely:

z₁ = cis(π/6) <--its good to get used to thinking in terms of pi. you can write cis(30), if you prefer.
z₂ = cis(7π/6) (or cis(150) which is "a full circle plus 90 degees (450 degrees) divided by 3")
z₃ = 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.
November 27th 2012, 08:58 PM
ImConfused

Re: Solve using de Moivre's theorem: z^3=i

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?
November 27th 2012, 09:03 PM
MarkFL

Re: Solve using de Moivre's theorem: z^3=i

Deveno has used the pi character "π", not the letter "n".
November 27th 2012, 09:09 PM
Deveno

Re: Solve using de Moivre's theorem: z^3=i

Quote:

Originally Posted by ImConfused

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?

no it is supposed to be the symbol for "pi", which does not look very good in the default sans serif font used on these forums. i have changed the font to make it more recognizable.
November 27th 2012, 09:26 PM
ImConfused

Re: Solve using de Moivre's theorem: z^3=i

Oh! That makes much more sense!

Thanks soo much for your help :)
January 12th 2013, 01:47 PM
HallsofIvy

Re: Solve using de Moivre's theorem: z^3=i

Quote:

Originally Posted by ImConfused

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

This is wrong. tan(0) is 0, not "1/0" which does not actually exist, though you can think of it as " $+\infty$ ". What is true is that as $\theta$ goes to $\pi/2$ , $\tan(\theta)$ increases without bound ("goes to infinity"). In any case, you should recognise that, marking "i" on the complex plane, the angle is $\pi/2$ , not 0.

Quote:

so z^3=1cis0

No, as pointed out above. $z^3= cis(\pi/2)$ . 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 " $\sqrt{0^2+ 1^2}= 1$ " to see that.

Quote:

z1=1cis0
z2=1cis120
z3=1cis240

I feel like something is wrong with these solutions. (Worried)
Those are the cube roots of 1, not i.
HELP please!