Proving ln(z^alpha) = alpha ln(z) for Complex Numbers

[DieCommie]

Homework Statement

Show that

$$ln(z^\alpha) = \alpha ln(z)$$

where 'z' and 'alpha' are complex.

Homework Equations

$$ln \alpha = ln r + i(\theta + 2*n*\pi)$$

The Attempt at a Solution

For the left hand side I have $$ln (z^\alpha) = ln [(r_{1}e^(i\theta_{1}))^(r_{2}e^(i\theta_{2}))]$$

and for the right $$\alpha ln(z) = \alpha [ln r + i(\theta + 2 n \pi)]$$


I have played with it for a bit but seem to be going in circles...
Any tips/hints would be greatly appreciated! Thank you!

 

[Kreizhn]

We know first of all that in the Complex Field, the exponential and logarithmic function are still natural inverses of one another. We define for the logarithmic function the principle branch to ensure analyticity.

let modulus(z) = |z| and argument(z) = arg(z)

We first show that exp is the inverse of log.

$$log z = log |z| + i.arg(z)$$, so that

$$e^{log z} = e^{log z}e^{i.arg(z)} = |z|e^{i.arg(z)}=z$$ by definition

Conversely, let z = x + iy

$$log(e^z)=log(e^{x+iy})$$
$$=log(e^x e^{iy})$$
$$=log |e^x e^{iy}| + i.arg(e^x e^{iy})$$
$$=log(e^x) + iy$$ where log(e^x) is the real logarithm
$$=x+iy = z$$

We then know that $$z^\alpha = e^{\alpha log(z)}$$

since we can expand
$$e^{\alpha log(z)} = \displaystyle\left( e^{log(z)} \right)^\alpha$$
$$=z^\alpha$$

Thus since $$z^\alpha = e^{\alpha log(z)}$$ then
$$log(z^\alpha) = log(e^{\alpha log(z)} ) = \alpha log(z)$$

as required.