A crazy ant is standing on the origin. It begins by walking 1 unit in the +x direction and then turns 60 degrees counterclockwise and walks 1/2 units in that direction. The ant then turns another 60 degrees and walks 1/3 units in that direction. The ant keeps doing this endlessly. What is the ant's final position ? #Method by Which you can solve
1) Cartesian Method
2) Polar Coordinate method

#Hint :- Use formula for displacement,
D = 1/n(cos∅+ isin∅)

Math question :)

The problem is amenable to some simple complex number analysis (you have to work in radian measure, note that 45∘=π4).

Each step can be characterised as a vector in the complex plane. The first step is 1=1ei0, the second is 12eiπ4, and the nth step is 1nei(n−1)π4. The final position is the infinite sum of all of these complex numbers.

Let z=eiπ4.

Then define a series sum S(z)=1+12z+13z2+14z3+...

Note that zS(z)=z+12z2+13z3+14z4+...

And now observe: (zS(z))′=1+z+z2+z3+z4+...=11−z, where convergence is assured for complex |z|<1.

By integrating and rearranging, we find, S(z)=−1zln(1−z). (The constant of integration is easily shown to be zero).

Now find |S(eiπ4)|, which is the distance of the final position of the ant from the origin.

|S(eiπ4)|=|−e−iπ4ln(1−eiπ4)|

Since |z1z2|=|z1||z2|, the above is equal to |−e−iπ4||ln(1−eiπ4)|=|ln(1−eiπ4)|

(since |−e−iπ4|=1)

To find |ln(1−eiπ4)|, we first express 1−eiπ4 in the form reiθ.

1−eiπ4=1−cosπ4−isinπ4

Now r=(1−cosπ4)2+(sinπ4)2−−−−−−−−−−−−−−−−−−√=2−2cosπ4−−−−−−−−−√=2−2–√−−−−−−√

and θ=arctansinπ41−cosπ4=arctan(2–√+1)

and ln(reiθ)=lnr+iθ, giving:

|ln(1−eiπ4)|=|12ln(2−2–√)+iarctan(2–√+1)|=[arctan(2–√+1)]2+14[ln(2−2–√)]2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=1.20806...

Explanation:

“Mystery is when the spectator knows less than the characters in the movie. Suspense is when the spectator knows more than the characters in the movie.” The fact that two terms often considered synonyms have actually two opposite definitions may surprise, but once you go deeper the difference is even enlightening.