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This question already has an answer here:

I have studied the Newton's method for real functions, and these are the conditions to find a function real root on $[a,b]$:


$$f(a) \ f(b)<0$$ $$f'(x)>0 \quad (\text{or} \quad f'(x)<0) \qquad \forall \in [a,b] $$


Convergence theorem:

$$\text{sign}(f' f'') = \text{sign}(x_0 -x^*)$$


Iterations:

$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \qquad x_n\in \mathbb{R}$$


How can I use Newton's method to find the complex roots of a polynomial (e.g. $z^2+1$) ?


I know a real coefficients polynomial of degree $p$ has $p$ different complex roots (for Fundamental theorem of algebra)


But, how can I verify the convergence of the Newton succession with an initial value $z_0$?

$$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)} \qquad z_n\in \mathbb{C}$$


Thanks!

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marked as duplicate by LutzL, Raskolnikov, Aqua, Siong Thye Goh, user223391 Nov 11 '17 at 18:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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you don't need newton's method. you can use muller's method which can find complex roots, doesn't require a derivate but you need 3 points. the method is awesome! basic requirements: knowledge on how to multiply, add, subtract, and divide. knowledge on how to take square roots if you only know that method that you need to have a polar form function, it will just waste your time. as with square roots, you can find all roots by solving a system of equations: we have a number called (a+bi) we know that t^2=(a+bi) 2xy=b x^2-y^2=a x^2+y^2=(a^2+b^2)^0.5 solve for equation 2 and 3. then find x and y. obviously, x is real and y is not. y=i x=x

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    $\begingroup$ This is not an answer, because the OP speciffically asks "How can I use Newton's method?" $\endgroup$ – fonini Nov 10 '17 at 23:58

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