Integer power of a complex number
Integer power of a complex number
Polar form
Let z be a complex number written under its polar format,
`z = r *( cos(\varphi) + i * sin(\varphi))`
n is a relative integer then, according to De Moivre's formula,
`z^n = r^n *( cos(n*\varphi) + i * sin(n*\varphi))`
Let z be a complex number expressed in its exponential form next,
`z = r * e^(i*\varphi)`
n is a relative integer then,
`z^n = r^n * e^(n*i*\varphi)`
This result is the same as seen above with De Moivre's formula.
Algebraic form
z is a complex number expressed in an algebraic form,
`z = x + i*y`, x and y are two real numbers
n is a relative integer then,
`z^n = (x + i*y)^n`
For small values of n, we can use remarkable identities, for example,
`z^2 = (x+i*y)^2 = x^2 -y^2 + i * (2*x*y)`
`z^3 = (x+i*y)^3 = x^3 -3*x*y^2 + i * (3*x^2*y-y^3)`
For large values of n, we may expand `(x + i*y)^n` by using the binomial coefficients but we recommend convert z from its algebraic form to its polar form and use De Moivre's formula (see above).
Example
Calculate `z^25` for `z = 1 - i`
We convert z to its polar form z = (`sqrt(2)`, `-pi/4`). We have, according to De Moivre's formula,
`z^25 = sqrt(2)^25 *( cos(-25*pi/4) + i * sin(-25*pi/4))`
`z^25 = sqrt(2)^25 *( cos(25*pi/4) - i * sin(25*pi/4))`
`z^25 = sqrt(2)^25 *( cos(6*pi +pi/4) - i * sin(6*pi+pi/4))`
`z^25 = sqrt(2)^25 *( cos(pi/4) - i * sin(pi/4))`
`z^25 = 4096*sqrt(2) *( sqrt(2)/2 - i * sqrt(2)/2)`
`z^25 = 4096 *( 1 - i )`
See also
Polar and exponential formats of a complex number
Algebraic form of a complex number