Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read some posts about the Mandelbrot. I read that the Mandelbrot should be defined by $f(z)=z^2+C$. In my understanding, I think, the $C$ should be a constant, like $0.27$ or $2.1+4.5i$. However, in some programming language source code, I found it.

temp <- temp * temp + current
iteration <- iteration + 1
goto Loop

I only paste the core calculation part. As you can see from here, the temp is the $z$, and current is $C$. However, the current is changing when it select different point, which means the $C$ is changing. I don't quite understand about it. From the equation, I thought the $C$ should be constant, why it's changing from point to another one. Could someone help me about it?

Best Regards,

share|improve this question
Well, of course $c$ has to change; from Wikipedia: "a complex number, $c$, is part of the Mandelbrot set if, when starting with $z_0 = 0$ and applying the iteration $z_{n+1} = z_n^2 + c$ repeatedly, the absolute value of $z_n$ never exceeds a certain number however large $n$ gets." You pick points $c$ in the complex plane and iterate... it stands to reason you need to change $c$. –  J. M. Sep 3 '11 at 11:45
add comment

1 Answer

up vote 1 down vote accepted

Check out the Wikipedia article, which vindicated the quoted program.

The Mandelbrot set is a $2$-dimensional section of a $4$-dimensional object, which is dependent on two complex quantities: starting point $z_0$ and $C$. In the Mandelbrot set, $z_0$ is fixed to zero (equivalently, $C$), and $C$ varies. Julia sets are the reverse, $C$ is held constant and $z_0$ varies.

share|improve this answer
Thanks. I think I got what your mean. Actually, there are two variable, Z and C. I just regard C as a const. However, the Z is const (0,0) in the Mandelbrot set. I mixed up something else. –  Yongwei Xing Sep 3 '11 at 12:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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