1
$\begingroup$

(1) $z_n=(\frac{1+i}{\sqrt{3}})^n$.

(2) $z_n=(\frac{1+i}{\sqrt{2}})^n$.

Why does (1) converge to 0 and (2) does not? Shouldn't they both diverge since $cos(\frac{n\pi}{4})$ and $sin(\frac{n\pi}{4})$ from $n=1$ to $\infty$ diverge.

(3) $h(z)=\frac{|z|}{z}, z\neq0$.

Idk what to do (3), it's been causing me pain. If I input large values of $z$ I get 1,-1,i,i.

  • 1
    Given $z\in\mathbb{C}$, do you know a sufficient condition for the sequence $(z^n)$ to converge to $0$ ? to diverge ? – 2017-01-29
  • 0
    Can I have an example? – 2017-01-29
  • 0
    You have to know that if $\vert z\vert<1$, then $\lim_{n\to\infty}z^n=0$. Also, if $\vert z\vert>1$ then the sequence $(z^n)_{n\ge 0}$ is unbounded and hence diverges. This doesn't give you all the information but it's a good starting point. – 2017-01-29
  • 0
    I see now. Thanks – 2017-01-29
  • 0
    Polar cordinates can help – 2017-01-29

1 Answers 1

0

The modulus of $1+i$ is $\sqrt{2}$, so the modulus of the terms in (2) is $1$ but the argument bounces around. In (1), the modulus of the terms is less than one, so even though the argument is all over the place, the terms get closer and closer to $0$. For (3), multiply top and bottom by the conjugate of $z$ and see if that helps.

  • 0
    for (3) I would get $\frac{x^3+xy^2+i(x^2y^2+y^4)}{x^2+y^2}$. Should I let z=x and z=ix and find the limits? – 2017-01-29
  • 1
    @HiPolyEraser Well, $z\overline{z} = |z|^2$, and one $|z|$ cancels. $h(z)$ lives on the unit circle. – 2017-01-29
  • 0
    @HiPolyEraser. (3) is an interesting question: if $h(z)$ is on the unit circle, what does than mean in terms of convergence/divergence? – 2017-01-29
  • 0
    @imranfat Yea (3) is causing me pain. Can't I represent h(z) as $h(z)=e^{-i\theta}$? It's to jumps to $\frac{\pi}{2}$ and $-\frac{\pi}{2}$. – 2017-01-29
  • 1
    @HiPolyEraser There's no $n$ in (3), so it's not really clear what one is supposed to do. – 2017-01-29
  • 0
    @HiPolyEraser. You already established that different values for $z$ yield outcomes that all have an absolute value of 1.But the complex numbers themselves are different. Do you think this possibly could be convergent? – 2017-01-29
  • 0
    @imranfat it does not converge – 2017-01-30
  • 0
    @HiPolyEraser. True – 2017-01-30