Cube root of i (complex numbers)?

Question

What is the cube root of i? there are three of em.
It has no real part, only imaginary part.
Thanks in advanced.

using the formula:
z=r(cos(theta)+isin(theta)).
r=i
but because there is no real part.
cosine suppose to be zero, so theta = pi/2
but in that case, i have i(0+i*1) and i get -1.
am i doing something wrong?

Answer 1

By far the easiest way to do this type of root problem is with the polar representation of the complex numer. That is put it in the form:

z = c + i*d

z = r * e^(i*a)

Where r = squareroot(c^2 + d^2)
a = arctan(d/c) with the proper quadrant implied


Where r and a are both real. The cube root of the real r is straightforward. Since the (i*a) is an exponents, taking the cube root just means dividing the angle by 3. The first root is then:

z^(1/3) = r^(1/3) e^(i*a/3)

Of course, in polar notation, increasing the angle by 2*pi yields the same number back again:

z = r * e^(i*(a +2*pi))

Again, the cube root uses the cube root of r and 1/3 the angle:

z^(1/3) = r^(1/3) * e^(i*(a +2*pi)/3) = r^(1/3) * e^(i*(a/3 +2*pi/3))

Adding 2*pi to the angle again, the cube root of z is:

z^(1/3) = r^(1/3) * e^(i*(a/3 + 4*pi/3))

If you try to do it again:

z^(1/3) = r^(1/3) * e^(i*(a/3 + 2*pi)) = r^(1/3) * e^(i*a/3)

You get the first root back again. So there are three roots. You might notice that all three roots are equally spaced around the full 2*pi circle. In fact, it is easy to show that all the roots for the nth root of z are:

For: z = r * e^(i*a)

z(1/n) = r^(1/n) * e^(i*(a/n + 2*pi*k/n))

for k ranging from 0 to n-1.

Hopefully this will let you do roots of complex numbers. Your original question was for the cube roots of i. For this:

z = i = r * e^(i*a) r = 1, a = pi/2

1. z(1/3) = r^(*1/3) e^(i*pi/2/3)
= cos(pi/6) + i*sin(pi/6)
= squareroot(3/4) + i/2
2. z(1/3) = r^(*1/3) e^(i*(pi/2 + 2*pi)/3)
= cos(5*pi/6) + i*sin(5*pi/6)
= -squareroot(3/4) + i/2
3. z(1/3) = r^(*1/3) e^(i*(pi/2 + 4*pi)/3)
= cos(3*pi/2) + i*sin(3*pi/2)
= 0 - i

Answer 2

Cube Roots Of I

Answer 3

The only cube root of i with no real part is -i. The other two are (i+√3)/2 and (i-√3)/2, which have real parts of √3/2 and -√3/2, respectively.

Edit: "am i doing something wrong?" Yes. You are doing two things wrong:
#1, r is always a positive real number. In particular, r≠i, r=1. Remember, r is the distance of the number from the origin of the complex plane, not the number itself.
#2: cos θ=0 does not imply θ=π/2. θ might also equal, say, 3π/2 (and does in this case).

Answer 4

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This involves putting your number into polar form. If you start in polar form, things are easy. If your original form is: z = x + i*y Then the polar form is: z = r*e^(i*a) Where: r = sqrt(x^2 + y^2), a = atan(y/x) Be careful with the atan. Make sure it is in the right quadrant. The roots of a complex number in polar form are: z ^ (1/n) = (r^(1/n)) * e^(i * ((a + 2*pi*k)/n)) Where k ranges from 0 to n-1, giving all n roots of z. If you need to convert from polar back to cartesian form: For z = r*e^(i*a) = x + I*y Where x = r*cos(a), y = r*sin(a)

Answer 5

You can do it the following way:
A + Bi = cuberoot(i)
Cube both sides
A^3 + 4a^2Bi + 4aB^2i^2 + B^3i^3 = 0 + i
Convert the powers of i into their simpler forms.
combine the i terms together and the real terms together
You'll get ( real terms) + (imaginary terms) i = 0 + i
for the two sides of this equation to be equal the 2 real parts have to be equal and the coefficients of i have to be equal. You'll get the 2 equations like the following:

real terms = 0
and
imaginary terms = 1

solve these 2 equations for all possible A and B and you'll have your answers.

I tried cubing Pascal's answer and it didn't come out right.
I finished it the way I explained and got the following answers:
-i
2/cuberoot(15) + (1/cuberoot(15) )i
2/cuberoot(15) - (1/cuberoot(15) )i

Answer 6

This Site Might Help You.

RE:
Cube root of i (complex numbers)?
What is the cube root of i? there are three of em.
It has no real part, only imaginary part.
Thanks in advanced.

Answer 7

we know e^(i*pi/2) = i as e^(2pi*i) = 1

now e^(ix) = cos x + i sin x
cube root of i = e ^(i*pi/6+i*2npi/6) = cos pi/6+ i sin pi/6
= 1/2+i(sqrt(3)/2)

n= 0 is above solution
n = 1 gives e^(i*(pi/6+pi/3) = e^i(pi/2) = cos pi/2+i sin pi/2 = -i
n = 2 gives e^(i(pi/6+2pi/3) = e^(i(5pi/6) = cos 5pi/6+ i sin pi/6
= 1/2 =i(ssqrt(3)/2



n = 2