6
$\begingroup$

I am trying to calculate $\exp z$ using $\ln z$ via Newton-Raphson method $$x_{n+1} = x_n-\frac{f(x_n)}{f^{'}(x_n)}$$and got the formula $$x_{n+1}=x_n-\frac{\ln x_n-z}{\frac{1}{x_n}}$$ where $z = a + bi$ is the value i'm trying to iterate to. However this formula only converges when $-3< b < 3$. But, if $z$ is instead any real number the formula converges accurately. Why does the formula not converge to any complex $z$, and how can I fix it so that it does?

$\endgroup$
  • 4
    $\begingroup$ It's almost certainly a problem with the branching of the complex log. $\endgroup$ – Randall Dec 2 '18 at 20:03
  • $\begingroup$ yes it means that b has to be less that 3 and greater than -3 to converge towards $e^z$ $\endgroup$ – user10560552 Dec 2 '18 at 20:03
  • $\begingroup$ Randall can you go more into detail about the branching of the complex log? $\endgroup$ – user10560552 Dec 2 '18 at 20:06
  • 1
    $\begingroup$ Not really, because it's been 20 years since I thought about this stuff intimately. You need someone with a better answer, who will certainly be along shortly. But, the complex exponential is not one-to-one, so what do you mean by "$\ln z$"? $\endgroup$ – Randall Dec 2 '18 at 20:08
  • $\begingroup$ The complex logarithm is $ln(z)$ and I know that $ln(z)$ is multivalued, along with $e^z$ like the trigonometric functions however, the formula should still be able to evaluate over all complex numbers right? $\endgroup$ – user10560552 Dec 2 '18 at 20:13
3
$\begingroup$

The complex, principal logarithm $\text{Log}(z)$ is defined by $$ \text{Log}(z) = \ln|z| + i \ \text{Arg}(z), $$ where $\text{Arg}$ stands for the principal argument of $z$ - that is $\text{Arg}(z)$ represents the angle that $z$ makes against the positive real axis when viewed in polar form. By the principal argument, we mean that $$-\pi < \text{Arg}(z) \leq \pi.$$ As a result, $\text{Log}(w) = z$ has no solution if $|\text{Im}(z)|>\pi$ so you cannot to expect to use that formulation to compute $e^z$.

To fix this, you can define your own branch of the logarithm with the appropriate value of it's imaginary part. For example, since 12 is between $3\pi$ and $5\pi$, you can define your logarithm by $$ \text{LOG}(z) = \ln|z| + i \ (\text{Arg}(z) + 4\pi). $$ Here's a simple implementation in Sage..

$\endgroup$
  • 1
    $\begingroup$ Thanks, that makes a lot of sense, it's like how arccosine needs to have a domain restriction, because otherwise it wouldn't be a function. $\endgroup$ – user10560552 Dec 3 '18 at 1:06
  • 1
    $\begingroup$ That's exactly right. The complex exponential is periodic with period $2\pi i$ so you can define a branch of the logarithm whose range lies in a strip of height $2\pi$. $\endgroup$ – Mark McClure Dec 3 '18 at 1:20
  • $\begingroup$ Just wondering why did you choose $4\pi$ wouldn't $5\pi$ also work? $\endgroup$ – user10560552 Dec 3 '18 at 1:40
  • 1
    $\begingroup$ We know that $-\pi<Arg(z)\leq \pi$ so I added on two periods of $2\pi i$ each to get $3\pi < Arg(z)\leq 5\pi$ which works since 12 is between these numbers. More generally, you want to add on some integer multiple of the period $2\pi i$ so i wouldn't expect $+5\pi$ to work. $\endgroup$ – Mark McClure Dec 3 '18 at 2:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.