Does the complex number i i belong to the Mandelbrot set?

Calculus Level 3

The Mandelbrot set is the set of complex numbers c c for which the function f c ( z ) = z 2 + c f_{c}(z)=z^{2}+c does not diverge when iterated from z = 0 , z=0, i.e., for which the sequence f c ( 0 ) , f c ( f c ( 0 ) ) , f c ( f c ( 0 ) ) , e t c , f_{c}(0), f_{c}(f_{c}(0)), f_{c}(f_{c}(0)), etc, remains bounded in absolute value. Whether a number in the complex plane lies in the Mandelbrot set or not depends on these two possibilities:

  1. When starting with z 0 = 0 z_{0} = 0 and applying the iteration repeatedly, the absolute value of z n z_{n} remains bounded however large n n gets. It's a property of the function that the upper bound never gets larger than 2 2 .
  2. When starting with z 0 = 0 z_{0} = 0 and applying the iteration repeatedly, the absolute value of z n z_{n} is unbounded such that for certain iterations the modulus of the value of the function exceeds 2. It so happens that the higher the number of iterations for such a complex number, the higher would be the modulus of the result.

This is best explained by an example. Let us find out whether 1 1 belongs to the Mandelbrot set We have: f c ( z ) = z 2 + c f_{c}(z)=z^{2}+c For the first iteration, we have z = 0 z=0 and c = 1 c=1 Thus f 1 ( 0 ) = 0 2 + 1 = 1 f_{1}(0)=0^{2}+1 = 1 For the next iteration, we put z = 1 z=1 and c = 1 c=1 Thus f 1 ( 1 ) = 1 2 + 1 = 2 f_{1}(1)=1^{2}+1 = 2 Next, we have f 1 ( 2 ) = 2 2 + 1 = 5 f_{1}(2)=2^{2}+1 = 5 Since the modulus of 5 5 is 5 5 which is greater than 2 2 , the number 1 1 does not belong to the Mandelbrot Set.

In contrast, when we take the complex number 1 -1 : For the first iteration, we have, f 1 ( 0 ) = 0 2 + ( 1 ) = 1 , f_{-1}(0)=0^{2}+ (-1) = -1, modulus of (-1) is 1 Next we have, f 1 ( 1 ) = ( 1 ) 2 + ( 1 ) = 0 , f_{-1}(-1)=(-1)^{2}+ (-1) = 0, modulus of (0) is 0 For the next iteration we have, f 1 ( 0 ) = 0 2 + ( 1 ) = 1 , f_{-1}(0)=0^{2}+ (-1) = -1, modulus of (-1) is 1 Hence, we keep on alternating between 1 1 and 0 0 and the values of never exceed 2.

Similarly, we can prove that 0 0 lies in the Mandelbrot set.

Can you tell whether i i lies in the Mandelbrot set or not?

The given image is a mathematician's depiction of the Mandelbrot set M. A point c c is colored black if it belongs to the set, and white if not. Re[c] and Im[c] denote the real and imaginary parts of c c , respectively.

Yes No

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1 solution

Pravya Agarwal
Mar 6, 2019

i i does indeed belong to the Mandelbrot set.

We have: f c ( z ) = z 2 + c f_{c}(z)=z^{2}+c

Thus, for the first iteration, f i ( 0 ) = 0 2 + i = i , i = 1 f_{i}(0)=0^{2}+i=i, |i|=1

Next, f i ( i ) = i 2 + i = i 1 , i 1 = 2 f_{i}(i)=i^{2}+i=i-1, |i-1|=\sqrt{2}

Next, f i ( i 1 ) = ( i 1 ) 2 + i = i , i = 1 f_{i}(i-1)=(i-1)^{2}+i=-i, |-i|=1

Thus, the values keep alternating between 1 1 and 2 \sqrt{2}

Here is a brilliant video by Numberphile explaining this concept: https://www.youtube.com/watch?v=NGMRB4O922I

Copy and paste this link in the URL bar to view the video.

Check again.........the values do not alternate. Try finding the next three iterates....

Aaghaz Mahajan - 2 years, 3 months ago

I did try the next three iterations and the values do seem to alternate.

Pravya Agarwal - 2 years, 3 months ago

The values start alternating between 1 + i -1 + i and i -i with the third term. As Pravya showed, f i ( 1 + i ) = i . f_i(-1+i) = -i. And it's easy to see f i ( i ) = 1 + i . f_i(-i) = -1+i.

Richard Desper - 2 years, 3 months ago

1 pending report

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