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:

yes its like krisnis like in the air force .

answer: #1 is (a) however

explanation: