# Newton-Raphson method for complex numbers

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.

• We can even use it for matrices although I can't quite explain why it works. – mathreadler May 13 '18 at 8:05
• If $A$ is a matrix can we find $Log_{ }A$ or something like that – Abhishek May 13 '18 at 8:07
• I have mostly tried polynomial because they often have nice convergence properties for Newton rhapsody – mathreadler May 13 '18 at 8:11
• Bless you automatic correct! – mathreadler May 13 '18 at 8:12
• @mathreadler we can find $e^A$ by using expansion of $e^x$ – Abhishek May 13 '18 at 8:14

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 $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0000000000+1.0000000000\, i \\ 1 & 0.1908866453+0.3494156605\, i \\ 2 & 0.3588322329+0.6585981708\, i \\ 3 & 0.3690703346+0.5754839470\, i \\ 4 & 0.3747254289+0.5764171936\, i \\ 5 & 0.3746990213+0.5764127231\, i \end{array} \right)$$

• Can we try it by substituting $x=a+ib$ – Abhishek May 13 '18 at 7:59
• @abhishekchaudhary. You can do it but then you will have to solve two equations for $a,b$. Newton-Raphson would work (making probably the problem more difficult). – Claude Leibovici May 13 '18 at 8:12
• But in your method iteration will be very difficult with complex numbers – Abhishek May 13 '18 at 8:13
• @abhishekchaudhary. You can solve $$e^a (a \cos (b)-b \sin (b))=0$$ $$e^a (a \sin (b)+b \cos (b))=1$$ – Claude Leibovici May 13 '18 at 8:18
• These are easily possible with scientific calculator but your one need some manual work – Abhishek May 13 '18 at 8:19

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

• Can you solve the above equation for me – Abhishek May 13 '18 at 7:43