Can Newton's Method be used to improve the accuracy of a complex root?

Question

Newton's method is known for not being able to find complex roots of a polynomial with real coefficients. But given that I have a complex initial approximation, will applying Newton's recursive formula to it: $$z_{n+1} = z_n - \frac{f(z_n)}{f'(z_n)}$$ result in a better approximation?

Answer 1

I did two examples. Looks pretty good to me. My original function was $z^2 + 1,$ no real roots. It was necessary to give it decimal starting value, otherwise it gives increasingly complicated rational expressions.

parisize = 4000000, primelimit = 500509
? I * I
%1 = -1
? 

? 

? z = 1.3 + 0.7 * I
%6 = 1.300000000000000000000000000 + 0.7000000000000000000000000000*I

? z = (z^2 - 1  ) / ( 2 * z )
%7 = 0.3518348623853211009174311927 + 0.5105504587155963302752293578*I
? z = (z^2 - 1  ) / ( 2 * z )
%8 = -0.2816651961787748391129345648 + 0.9192771384378602102396311448*I
? z = (z^2 - 1  ) / ( 2 * z )
%9 = 0.01151673510543323839055292482 + 0.9568645700250763626307322269*I
? z = (z^2 - 1  ) / ( 2 * z )
%10 = -0.0005299653500082591013398663256 + 1.000896586227707046035537310*I
? z = (z^2 - 1  ) / ( 2 * z )
%11 = -4.745955222464034310471682037 E-7 + 1.000000261518840743450392327*I
? z = (z^2 - 1  ) / ( 2 * z )
%12 = -1.241156755611569871523718596 E-13 + 0.9999999999999215756765779129*I
? z = (z^2 - 1  ) / ( 2 * z )
%13 = 9.733687881959195813 E-27 + 0.9999999999999999999999999954*I
? z = (z^2 - 1  ) / ( 2 * z )
%14 = 0.E-44 + 1.000000000000000000000000000*I
? 


Start Over :

? z = 57.6 -  74.2 * I
%15 = 57.60000000000000000000000000 - 74.20000000000000000000000000*I
? z = (z^2 - 1  ) / ( 2 * z )
%16 = 28.79673595212729786703538319 - 37.10420472833601559489539180*I
? z = (z^2 - 1  ) / ( 2 * z )
%17 = 14.39184098711877879304227246 - 18.56051230113005287001384541*I
? z = (z^2 - 1  ) / ( 2 * z )
%18 = 7.182875371107810147135708494 - 9.297079860328703953240807638*I
? z = (z^2 - 1  ) / ( 2 * z )
%19 = 3.565418316636721133579796819 - 4.682217829089264656490428861*I
? z = (z^2 - 1  ) / ( 2 * z )
%20 = 1.731238390793015004421757818 - 2.408701921586626731300483351*I
? z = (z^2 - 1  ) / ( 2 * z )
%21 = 0.7672425486244663387358949699 - 1.341224091252557585696157024*I
? z = (z^2 - 1  ) / ( 2 * z )
%22 = 0.2229451132702457867657008095 - 0.9514915907790448803050888252*I
? z = (z^2 - 1  ) / ( 2 * z )
%23 = -0.005247650540332528288061227377 - 0.9738876652533082699217210838*I
? z = (z^2 - 1  ) / ( 2 * z )
%24 = 0.0001425084495941948420301598225 - 1.000335162125949610138930824*I
? z = (z^2 - 1  ) / ( 2 * z )
%25 = 4.774087805583942772563298163 E-8 - 1.000000046003880952632565108*I
? z = (z^2 - 1  ) / ( 2 * z )
%26 = 2.196265573505168302989229217 E-15 - 0.9999999999999999185829211779*I
? z = (z^2 - 1  ) / ( 2 * z )
%27 = -1.7881352731234044628 E-31 - 1.000000000000000000000000000*I
? 

Answer 2

First, let's try to clarify your question by referring your comment:

My professor said that applying a real number to Newton's method will only give you real approximations.

I feel certain that the correct statement is:

Applying Newton's method to a real initial seed to will only give you real approximations.

This is absolute true. After all, if $f$ is a polynomial with real coefficients then $$N(z) = z - \frac{f(z)}{f'(z)}$$ maps $\mathbb R \to \mathbb R \cup \{\infty\}$. Thus, $N$ cannot produce complex numbers so it cannot produce a sequence that converges to a complex root.

---

Now, you ask:

will applying Newton's recursive formula ... result in a better approximation?

So, I guess the question is - even though Newton's method cannot produce a sequence that converges to a complex root, will it least move the terms closer and closer to the root?

The answer to this question is no and we can establish that easily enough by example. Consider $f(x)=x^2+3$. Then, $$N(x) = x - \frac{x^2+3}{2x}$$ and it's easy to show by direct computation that $N(1)=-1$ and $N(-1)=1$. Thus, we have a two cycle with neither point moving any closer to the root.

Much more is true, though. In fact, the real line forms the Julia set in the complex plane for the function $N$ and the dynamics of $N$ are chaotic there. If you have a computer, you might try iterating $N$ from a randomly chosen real number and plotting the points on a line. I did so and generated the following image:

We can see that, far from moving the points closer to the root, they seem to jump around all over the place.