Finding the smallest non-negative integer $n$ such that $z_n=\left(1+\frac{i}{10}\right)^n$ lies in the Second Quadrant

Question

I am in a high school precalc/calc ab class. I would really appreciate it if someone could help me solve this homework problem. I'm totally lost!

Consider the sequence of complex numbers $z_n=\left(1+\frac{i}{10}\right)^n$. It is understood that $n$ stands for a nonnegative integer. Notice that $z_1$ and $z_2$ are in the first quadrant. What is the smallest positive value of $n$ for which $z_n$ is in the second quadrant?

We have not covered arguments, derivatives, or anything complex of that sort (if those were needed). I understand that in order for $z_n$ to be in the second quadrant, the real portion must be negative, but I don't know how to find the correct $n$ without guess-and-check (which would become really messy).

The answer to the question is $n=16$. Could someone explain how to get there using basic complex number principles?

Edit: Thank you so much!

Answer 1

Hint Converting to polar coordinates gives that $$1 + \frac{i}{10} = \frac{\sqrt{101}}{10} e^{i \arctan \frac{1}{10}},$$ so, geometrically, multiplying by $1 + \frac{i}{10}$ is a dilation by a factor of $\frac{\sqrt{101}}{10}$ and a rotation by $\arctan \frac{1}{10} = \operatorname{arccot} 10$.

So, how many times $n$ do we need to multiply by $1 + \frac{i}{10}$ for the total rotation to be more than $\frac{\pi}{2}$?

$\left\lceil\frac{\frac{\pi}{2}}{\operatorname{arccot} 10}\right\rceil = 16$

Answer 2

The argument of a complex number is the angle between the segment from the origin to the number and the $x$ axis. You can compute the argument of $1+\frac i{10}$ with an $\arctan$. When you multiply complex numbers you add their arguments, so you need to find the $n$ you multiply the argument you found by to get over $\frac \pi 2$ to get into the second quadrant.