find all complex solutions of X^5 - i = 1?

Answer 1

x^5 - i = 1

x^5 = 1 + i

x^5 = (√2)*(1/√2 + i/√2)

x^5 = 2^(1/2) * (cos 45 + i*sin 45)

The above is the simplified Euler notation for our problem, now we can go ahead and find the roots:

x = 2^(1/10) * {(cos 45 + i*sin 45)^(1/5)}

Using Euler's Formula identity,

x = 2^(1/10) * {cos (45/5) + i*sin (45/5)}
x = 2^(1/10) * (cos 9 + i*sin 9)

this is one root of the eqation; to get the other roots, rewrite the original angle of 45 degrees by adding 360 degrees four times (the fifth time, it will cycle back to your first solution). Remember that by adding 360 degrees to any angle will not alter the result or properties of the angle in any way.

Second root:
x = 2^(1/10) * [ {cos (45+360) + i*sin (45+360)} ^ (1/5) ]
x = 2^(1/10) * { cos (405/5) + i*sin (405/5) }
x = 2^(1/10) * { cos 81 + i*sin 81 }

Third root:
x = 2^(1/10) * [ {cos (405+360) + i*sin (405+360)} ^ (1/5) ]
x = 2^(1/10) * { cos (765/5) + i*sin (765/5) }
x = 2^(1/10) * { cos 153 + i*sin 153 }

Fourth root:
x = 2^(1/10) * [ {cos (765+360) + i*sin (765+360)} ^ (1/5) ]
x = 2^(1/10) * { cos (1125/5) + i*sin (1125/5) }
x = 2^(1/10) * { cos 225 + i*sin 225 }

Fifth root:
x = 2^(1/10) * [ {cos (1125+360) + i*sin (1125+360)} ^ (1/5) ]
x = 2^(1/10) * { cos (1485/5) + i*sin (1485/5) }
x = 2^(1/10) * { cos 297 + i*sin 297 }

Sixth root (possible?):
x = 2^(1/10) * [ {cos (1485+360) + i*sin (1485+360)} ^ (1/5) ]
x = 2^(1/10) * { cos (1845/5) + i*sin (1845/5) }
x = 2^(1/10) * { cos 369 + i*sin 369 }
but note that this root can be simplified as:
x = 2^(1/10) * { cos (360+9) + i*sin (360+9) }
x = 2^(1/10) * (cos 9 + i*sin 9)
which is the same as our first root.


Therefore the five roots or complex solutions are:
(a) x = 2^(1/10) * ( cos 9 + i*sin 9 )
(b) x = 2^(1/10) * { cos 81 + i*sin 81 }
(c) x = 2^(1/10) * { cos 153 + i*sin 153 }
(d) x = 2^(1/10) * { cos 225 + i*sin 225 }
(e) x = 2^(1/10) * { cos 297 + i*sin 297 }

Note: the angles are in degrees, you can use a scientific calculator to determine 2^(1/10) and the sines & cosines of the above angles. This will give the numerical values of the solutions.

Answer 2

That's i as in the complex number, I'm assuming.

Well the easiest way is if you know the rule for roots in complex numbers. You take the absolute value (distance from the origin), take the Nth root of that, and then take the Nth fraction of your angle. So here we have x^5 = 1+i. 1+i is a point with an angle of 45 degrees and an absolute value of square-root-2. So the fifth root of that is a point with an angle of 15 degrees and an absolute value of fifth-root-of-square-root-2, or 2^(1/10).

So x is the point 2^(1/10) angle 15 degrees. To convert that back, use sin and cosine (CIS): x = 2^(1/10) cis 15deg = 2^(1/10) (cos 15 + i sin 15). This cannot be simplified but you can work it out with a calculator.

If you are looking for the other four roots (since this is a fifth-degree problem), just rotate that angle five times around a circle. Since 360/5 = 72, we have the original root at 15 degrees, then 87, then 159, then 231, then 303, then 375 - which is just 360 plus 15, and our original position again.

Answer 3

Here is a list of the solutions on a pdf

http://www.geocities.com/euleratuo/