3

This is on my homework on differentials and partial differentiation, so I'm not sure what application these could have on the natural log of z

stack ex
  • 350
  • 1
  • 3
  • 16

2 Answers2

12

Let $z=x+iy$. To find real and imaginary part of $\ln(x+iy)$ let $x+iy=re^{i\theta}$ where $r= \sqrt{x^2+y^2}$ and $\theta=\tan^{-1}{\frac{y}{x}}$ therefore, $$\ln(x+iy)=\ln\left(re^{i\theta}\right)=\ln(r)+i\theta \ln(e)= \ln(r)+i\theta$$ Hence, real part $$\ln(r)$$ i.e $$\ln\left(x^2+y^2\right)^{1/2}= \frac{1}{2}\ln\left(x^2+y^2\right)$$ and imaginary part is $$\theta=\tan^{-1}\frac{y}{x}$$

Notice here $$\ln(x+iy)= \ln(r)+ i\theta= \frac{1}{2}\ln\left(x^2+y^2\right)+i\tan^{-1}\frac{y}{x}$$

Community
  • 1
Iti Shree
  • 1,506
0

The answer of Iti Shree is correct, but under tacit assumptions, which I would like to clarify here. In particular, $\Im(\ln(z)) = \arctan(\frac{y}{x})$ is not defined if $x = 0$ and does not distinguish between opposite complex numbers.

First, $\ln(z)$ needs to be defined properly. Notice that it cannot be defined continuously on the whole complex plane, because of the branch point at $z = 0$. But it can be defined on the complex plane without the negative real numbers as the unique analytical continuation of the standard logarithm defined on the positive real numbers. This definition is assumed here.

Also, $\arctan(x)$ needs to be defined properly. Let $\arctan(x)$ be defined such that $-\frac{\pi}{2} < \arctan(x) < \frac{\pi}{2}$.

Finally, let $x$ denote $\Re(z)$ and $y$ denote $\Im(z)$.

Then, for $x > 0$ or $y \ne 0$:

$ \begin{aligned}[t] \Re(\ln(z)) &= \ln(|z|) = \frac{\ln(x^2 + y^2)}{2} \\ \Im(\ln(z)) &= \arg(z) = 2 \cdot \arctan\left(\frac{y}{x + \sqrt{x^2 + y^2}}\right) \end{aligned} $

where $\arg(z)$ is defined such that $-\pi < \arg(z) \le \pi$.

Loic
  • 821