How to calculate the argument and its limit for the sequence $z_n=-2+i\frac{(-1)^n}{n^2}$

Question

I am trying to show that the limit of the sequence $$z_n=-2+i\frac{(-1)^n}{n^2}$$ exists, using the polar representation. Note that $\lim_{n\rightarrow \infty }z_n=-2$. $$$$I am finding difficulty in arriving at the final answer (specifically in finding the limit of the argument - I am getting a wrong result)

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In polar form: $$ r=\sqrt{4+\frac{1}{n^4}}$$ $$\theta_0 =tan^{-1}(\frac{(-1)^n}{-2n^2})=-tan^{-1}(\frac{(-1)^n}{2n^2})$$

where $\theta_0$ is the principal argument defined on $-\pi <\theta _0\le \pi $.

Now taking the limit of the modulus, we get $$\lim_{n\rightarrow \infty }r=\sqrt{4+0}=2$$ and for the argument, we consider even and odd integers:

• If n is even:

• If n is odd:

But in the text book, the answers are

What did I do wrong? Isn't the inverse of an angle of $0$ is just $0$, which is is also in the range of the principal argument as in the figure below?

Moreover, in another example in the textbook (same sequence but with a positive real part) $$z_n=2+i\frac{(-1)^n}{n^2}$$ the limits of the argument were both $0$ and hence it exists. What is the difference here? Why both $0$ and not, say, $\pi$, as above?

Answer 1

Your statement $$\theta_0=\tan^{-1}\Bigl(\frac{(-1)^n}{-2n^2}\Bigr) =-\tan^{-1}\Bigl(\frac{(-1)^n}{2n^2}\Bigr)$$ is wrong.

In the case $n$ is even, this says $$\theta_0 =-\tan^{-1}\Bigl(\frac1{2n^2}\Bigr)\ .$$ Remember that $\tan^{-1}$ of a positive number gives an angle in the first quadrant, and so your answer is in the fourth quadrant. However, $z_n$ has negative real part and positive imaginary part, so it is in the second quadrant, so the argument is $$\pi-\tan^{-1}\Bigl(\frac1{2n^2}\Bigr)\ .$$

See if you can finish this off and then do the case of odd $n$ yourself.

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For the problem you mentioned at the end of the question, $z$ has positive real part and therefore is in the first or fourth quadrant. That's why the sort of calculations you did work out correctly.

Answer 2

Using $\theta_0$ as the principal argument of $z_n$ with $-\pi < \theta_0 \leq \pi$, it seems to me that when $n$ is even, we have

$$\tan\theta_0 = \frac{\mathrm{Im }\ z}{\mathrm{Re }\ z} = \frac{-1}{2n^2}$$ so that $$\theta_0 = \pi - \tan^{-1}\frac{-1}{n^2},$$

while for odd $n$, we have

$$\theta_0 = \tan^{-1}\left(\frac{1}{n^2}\right) - \pi.$$

If you really want, I can provide the crude drawings that lead me to believe this, but it hinges on the fact that $\tan^{-1}\theta \in (-\pi/2, \pi/2)$, while we need the principal argument $\theta_0$ to be between $-\pi$ and $\pi$.

If you look at what your angles are converging to, you'll see $0$. But that certainly shouldn't be the case, since $z_n$ converges to a value on the negative $x$-axis, where the angle is $\pi$. That should send up a red flag.