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!

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

I think I would write:

where hence:

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 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.

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?

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

**Deveno** has used the pi character "π", not the letter "n".

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.

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

Oh! That makes much more sense!

Thanks soo much for your help :)

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 " ". 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.

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!