Existence of sequence
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Existence of sequence
Show that there exists a sequence \( \displaystyle \left( z_{n} \right)_{n \in \mathbb{N}} \) of complex numbers convergent to 0 such that \( \displaystyle \lim_{n} {e}^{\frac{1}{z_{n}}} = 3 + 7i \).
Re: Existence of sequence
I think that this leads to a solution:
We substitute \(\frac{1}{z_n}=\theta_n\), and then it is enough to find a sequence \(\theta_n\) such that \(|\theta_{n+1}|>|\theta_n|+1\), such that \(e^{\theta_n}=3+7i\) for every \(n\). Since we are talking about complex numbers and by the periodicity of \(\sin,\cos\), we can find such a sequence, and then obviously \(z_n\rightarrow 0\)
We substitute \(\frac{1}{z_n}=\theta_n\), and then it is enough to find a sequence \(\theta_n\) such that \(|\theta_{n+1}|>|\theta_n|+1\), such that \(e^{\theta_n}=3+7i\) for every \(n\). Since we are talking about complex numbers and by the periodicity of \(\sin,\cos\), we can find such a sequence, and then obviously \(z_n\rightarrow 0\)
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- Community Team
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Re: Existence of sequence
I can't really tell for sure whether your arguments lead to a solution, but I think that they do. If I am not mistaken, you mean the following:
For each \( \displaystyle n \in \mathbb{N} \) define the set
\[ \displaystyle A_{n} = \left\{ x+iy \Big| 2(n-1)\pi \leq y < 2n\pi \right\} \]We know that \( \displaystyle {e}^{z} , z \in \mathbb{C} \) restricted to \( \displaystyle A_{n} \) is onto \( \displaystyle \mathbb{C} \smallsetminus \{0\} \), so for each \( \displaystyle n \in \mathbb{N} \) there is a \( \displaystyle \theta_{n} \in \mathbb{C} \) such that \( \displaystyle {e}^{\theta_{n}} = 3 + 7i \). Hence \( \displaystyle z_{n} := \frac{1}{\theta_{n}} \) has the desired properties.
Am I right?
What I had in mind is the following:
Consider the function \( \displaystyle f(z)={e}^{\frac{1}{z}} , z \in \mathbb{C} \smallsetminus \{0\} \) and observe that it is analytic in that region. So \( \displaystyle f \) can be expanded in a Laurent series there and we easily see that
\[ \displaystyle f(z) = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots \]This means that \( \displaystyle z_{0}=0 \) is an essential singularity of \( \displaystyle f \). If \( \displaystyle w = 3 + 7i \), by Casorati-Weierstrass theorem we have that there is a sequence \( \displaystyle \left( z_{n} \right)_{n \in \mathbb{N}} \subset \mathbb{C} \) with \( \displaystyle z_{n} \rightarrow 0 \) such that \( \displaystyle f(z_{n}) = {e}^{\frac{1}{z_{n}}} \rightarrow w = 3 + 7i \).
For each \( \displaystyle n \in \mathbb{N} \) define the set
\[ \displaystyle A_{n} = \left\{ x+iy \Big| 2(n-1)\pi \leq y < 2n\pi \right\} \]We know that \( \displaystyle {e}^{z} , z \in \mathbb{C} \) restricted to \( \displaystyle A_{n} \) is onto \( \displaystyle \mathbb{C} \smallsetminus \{0\} \), so for each \( \displaystyle n \in \mathbb{N} \) there is a \( \displaystyle \theta_{n} \in \mathbb{C} \) such that \( \displaystyle {e}^{\theta_{n}} = 3 + 7i \). Hence \( \displaystyle z_{n} := \frac{1}{\theta_{n}} \) has the desired properties.
Am I right?
What I had in mind is the following:
Consider the function \( \displaystyle f(z)={e}^{\frac{1}{z}} , z \in \mathbb{C} \smallsetminus \{0\} \) and observe that it is analytic in that region. So \( \displaystyle f \) can be expanded in a Laurent series there and we easily see that
\[ \displaystyle f(z) = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots \]This means that \( \displaystyle z_{0}=0 \) is an essential singularity of \( \displaystyle f \). If \( \displaystyle w = 3 + 7i \), by Casorati-Weierstrass theorem we have that there is a sequence \( \displaystyle \left( z_{n} \right)_{n \in \mathbb{N}} \subset \mathbb{C} \) with \( \displaystyle z_{n} \rightarrow 0 \) such that \( \displaystyle f(z_{n}) = {e}^{\frac{1}{z_{n}}} \rightarrow w = 3 + 7i \).
Re: Existence of sequence
Yes, this is what I meant.
Although I have to say I like your solution a lot more!
Although I have to say I like your solution a lot more!
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