2.5 Complex numbers

If z=r(cosθ+i sinθ), then z^n=r^n(cos nθ+i sin nθ). The theorem also gives a systematic route to trigonometric identities and roots of complex numbers.

Convert to modulus–argument form, multiply the argument by n and convert back only at the end. Arguments are defined modulo 2π, so equivalent angles represent the same complex number.

(cosθ+i sinθ)^3=cos3θ+i sin3θ. Equating real parts gives cos3θ=4cos³θ−3cosθ.

The modulus is raised to n as well as the angle; multiplying only the angle gives the wrong magnitude.

To solve w^n=z=r(cosθ+i sinθ), the roots have modulus r^(1/n) and arguments (θ+2πk)/n for k=0,1,…,n−1.

The n values are equally spaced on a circle and are distinct modulo 2π. Give them in polar form or convert each to Cartesian form as requested.

The cube roots of 8 are modulus 2 with arguments 0, 2π/3 and 4π/3, giving 2, −1+i√3 and −1−i√3.

Using only θ/n misses the other roots; the 2πk term is what enumerates the full set.

Expand (cosθ+i sinθ)^n using the binomial theorem, then equate its real and imaginary parts with cos nθ+i sin nθ to obtain trigonometric identities.

Separate even powers of i for the real part and odd powers for the imaginary part. Keep the combinatorial coefficients and signs organised before simplifying.

For n=2, the real part of (cosθ+i sinθ)^2 gives cos2θ=cos²θ−sin²θ, while the imaginary part gives sin2θ=2sinθcosθ.

The identity follows from equality of complex numbers, not from treating i as an ordinary positive number.