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From Wunsch Complex Analysis 5.2, 1 From home, you walk one mile east, turn 90° and walk 1/2 mile north, then turn 90° and walk 1/4 mile west, then turn 90° and walk 1/8 mile south, then turn 90° and go 1/16 th mile east. You continue on this journey, always turning 90° counterclockwise and making each segment of your trip equal to half the length of the previous one. There are an infinite number of such segments in the journey. If you plot your spiral voyage in the complex plane you can make each segment correspond to the terms in the infinite series $z = i/2$. A: When you have completed your trip what is the straight line distance, in miles, between your destination and your home?

MY Attempt: $\frac{1}{1+1/4}=4/5=x$ dist; $\frac{1}{2}\frac{1}{1+1/4}=2/5=y$ dist;

Total:$$=\sqrt{x^2+y^2}=\frac{\sqrt{20}}{{5}}$$

What's the intuition behind the author's method: $$\frac{1}{|1-i/2|}=\frac{2}{\sqrt{5}}$$ i understand how to mathematically simplify, but what's the reason for this?

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    $\begingroup$ Very nice problem! Your solution is correct, of course. But you was lucky that each time you turn 90°. It would be much harder if each time you turn f.e. 60°. Solution with complex numbers would remain as simple and elegant as it is for 90°. $\endgroup$
    – lesnik
    Aug 18, 2016 at 12:18

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The first number in this series, as a number in the complex plane, can be thought of as $1$ (that is, the number of length $1$ going "east"). Each successive path is half of the previous distance, rotated $90^o$ counterclockwise. This is equivalent to multiplying by $\frac{i}{2}$. Therefore, the limit of this sequence is given by $$\sum_{k=0}^\infty \left(\frac{i}{2}\right)^k = \frac{1}{1-\frac{i}{2}} = \frac{4}{5}+\frac{2}{5}i$$ And the absolute value of this number is $\frac{2}{\sqrt{5}}$. This is similar to real geometric series, such as $$\sum_{k=0}^\infty \left(\frac{1}{2}\right)^k = 1+\frac{1}{2}+\frac{1}{4} + ... = \frac{1}{1-\frac{1}{2}} = 2$$

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  • $\begingroup$ Ah, it makes sense now. I only have a scrawled out version to compare my attempt with. Using i you can sum the series in one step. Didn't think about that. :) $\endgroup$ Aug 18, 2016 at 11:38
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Consider home is at origin. Clearly x co-ordinate changes in every odd step and y co-ordinate changes in every even step. Considering the end positions in 1st few steps...

1st- (1,0)

2nd- (1,1/2)

3rd- (1-1/4 ,1/2)

4th- (1-1/4 ,1/2 - 1/8)

...

...

n->infinity - $((\frac{1}{1+\frac{1}{4}})$ , $\frac12(\frac1{1+\frac14}))$ (remember the sum of infinite G.P series)

now just calculate distance from origin of this point....i.e. $\frac{2}{\sqrt5}$

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