### Author Topic: Section 1.4 Question 1  (Read 1398 times)

#### Nikita Dua

• Jr. Member
•  • Posts: 14
• Karma: 0 ##### Section 1.4 Question 1
« on: October 01, 2018, 08:02:20 PM »
lim n -> infinity zn = ((1 +i)/(sqrt(3))n.
I used polar coordinates to solving giving r = sqrt(2/3)
xn =sqrt(2/3) cos(n * pi /4)
yn =sqrt(2/3) sin(n * pi /4)
zn = sqrt(2/3) cos(n * pi /4)  + isqrt(2/3) sin(n * pi /4)
From this how can show that it converges to 0?

#### oighea ##### Re: Section 1.4 Question 1
« Reply #1 on: October 02, 2018, 01:27:44 AM »
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.
« Last Edit: October 02, 2018, 01:41:18 AM by oighea »

#### Victor Ivrii ##### Re: Section 1.4 Question 1
« Reply #2 on: October 02, 2018, 09:54:22 AM »
In this problem argument is irrelevant, only modulus matters.

#### Min Gyu Woo

• Full Member
•   • Posts: 29
• Karma: 12 ##### Re: Section 1.4 Question 1
« Reply #3 on: October 02, 2018, 02:27:13 PM »
"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?

#### Victor Ivrii ##### Re: Section 1.4 Question 1
« Reply #4 on: October 02, 2018, 02:40:42 PM »
Look at the definition of $z_n\to A$ .