Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Nikita Dua on October 01, 2018, 08:02:20 PM

lim n > infinity z_{n} = ((1 +i)/(sqrt(3))^{n}.
I used polar coordinates to solving giving r = sqrt(2/3)
x_{n} =sqrt(2/3) cos(n * pi /4)
y_{n} =sqrt(2/3) sin(n * pi /4)
z_{n} = sqrt(2/3) cos(n * pi /4) + isqrt(2/3) sin(n * pi /4)
From this how can show that it converges to 0?

This sequence converges to 0 because the absolute value converges to 0. Note ${\sqrt{\frac{2}{3}}} < 1$, so powers of ${\sqrt{\frac{2}{3}}}^n$
Proof:
The Arg of $\frac{(1+i)}{\sqrt{3}}$ is $\frac{\pi}{4}$. That can be verified as $(1+i)$ "points northeast", and the $\sqrt{3}$ denominator is irrelevant to the Arg.
The magnitude $\frac{(1+i)}{\sqrt{3}}$ is ${\sqrt{\frac{2}{3}}}$. That can be verified as $1+i$ = $\sqrt2$, and $\frac{\sqrt2}{\sqrt3}$ = ${\sqrt{\frac{2}{3}}}$.
Therefore, $\frac{(1+i)}{\sqrt{3}} = {\sqrt{\frac{2}{3}}}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$
And by DeMoivre's law, $[\frac{(1+i)}{\sqrt{3}}]^n$ = ${\sqrt{\frac{2}{3}}}^n(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4})$
Intuitively magnitude of $[\frac{(1+i)}{\sqrt{3}}]^n$ can only spiral down counterclockwise as $n$ increases, and eventually approaches 0.

In this problem argument is irrelevant, only modulus matters.

"This sequence converges to 0 because the absolute value converges to 0."
Where did you get that information?
The textbook states on pg 34 that,
If $z_n \rightarrow A$ then $z_n \rightarrow A$.
Does the converse work as well?

Look at the definition of $z_n\to A$ .