Newton-Raphson method for complex numbers

Question

The Newton-Raphson Method as we know it is
$$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}$$ Where $x$ is solution of $f\left(x\right)=0$
But What if we have a equation of the form $$xe^x=i$$
Can we apply Newton-Raphson method treating $i$ as constant or we have to substitute $x=a+ib$ and solve two simultaneous equations.
Can you help me please?

Answer 1

As Robert Israel answered, the work to be done is the same and the same problems are faced, in particular the choice of $x_0$.

Using $x_0=1+i$, the first iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.0000000000+1.0000000000\, i \\ 1 & 0.5635771723+0.8175944690\, i \\ 2 & 0.3765837446+0.6414532983\, i \\ 3 & 0.3713945744+0.5772934579\, i \\ 4 & 0.3747064835+0.5764066745\, i \\ 5 & 0.3746990207+0.5764127230\, i \end{array} \right)$$ Using $x_0=1$, the first iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.0000000000+0.0000000000\, i \\ 1 & 0.5000000000+0.1839397206\, i \\ 2 & 0.2776785639+0.4861083567\, i \\ 3 & 0.3775795796+0.5918579469\, i \\ 4 & 0.3745261046+0.5765158277\, i \\ 5 & 0.3746990317+0.5764126917\, i \end{array} \right)$$ Using $x_0=i$, the first iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0000000000+1.0000000000\, i \\ 1 & 0.1908866453+0.3494156605\, i \\ 2 & 0.3588322329+0.6585981708\, i \\ 3 & 0.3690703346+0.5754839470\, i \\ 4 & 0.3747254289+0.5764171936\, i \\ 5 & 0.3746990213+0.5764127231\, i \end{array} \right)$$

Answer 2

Newton-Raphson is exactly the same for equations involving complex numbers. You just have to do the arithmetic using complex numbers.