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Let $\{z_n\}$ be a sequence of complex numbers, and suppose $\sum\limits_{k=0}^\infty z_k$ converges. Suppose $\exists \delta>0$, with $\delta < \pi/2$ such that $|\arg(z_k)|\le \delta$ for all $k\ge 0$. Prove that $\sum\limits_{k=0}^\infty z_k$ converges absolutely.

Proof:

Let $Z:= \sum\limits_{k=0}^\infty z_k$. Since $\sum\limits_{k=0}^\infty z_k$ converges, $z_k$ also converges, so let $z:=\lim\limits_{k\to \infty}z_k$. This implies that $\forall\varepsilon > 0, \exists N_1>0$ such that $|z_k-z|<\varepsilon$ whenever $k>N$. Also, $\exists N2$ such that $k>N_2$ implies $\left| \sum\limits_{k=0}^\infty z_k -Z\right|<\varepsilon$. Let $N=\max\{N_1, N_2\}$. Let $z_k = r_k e^{i\theta_k}$ and $\sum\limits_{k=0}^N z_k=se^{i\gamma_k}$, then

$$\left|\sum\limits_{k=0}^n |r_k e^{i\theta_k}|-|se^{i\gamma}|\right|...$$

But what next? What is so special about the absolute value of the argument of $z_k$ being less than $\pi/2$? Please help.

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  • $\begingroup$ Note that this is not true if $\delta = \pi/2$. For example take $z_k = 2^{-k}+(-1)^ki$. The fact that $\delta$ is strictly less than $\pi/2$ is absolutely essential for this to be true. $\endgroup$ – Shalop Feb 17 '17 at 3:59
  • $\begingroup$ Sorry, what should I note about $2^{-k}+(-1)^k i$? I see that its argument is $\pi/2$, but what follows from this? $\endgroup$ – sequence Feb 17 '17 at 4:12
  • $\begingroup$ It is true that $\sum z_k$ converges conditionally, but not absolutely, even though $|\arg z_k| < \pi/2$ for all $k$. $\endgroup$ – Shalop Feb 17 '17 at 4:15
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Since $z \mapsto Re(z)$ is continuous it follows that $\sum Re(z_k)$ also converges. Note that $Re(z_k)>0$ for all $k$, and thus $\sum Re(z_k)$ converges absolutely.

Now, using the given information we see that $|Im(z_k)/Re(z_k)| \leq \tan \delta$, therefore $|Im(z_k)| \leq \tan \delta |Re(z_k)|$. So by the comparison test $\sum Im (z_k)$ also converges absolutely.

Now using the fact that $|z_k| \leq |Im(z_k)|+|Re(z_k)|$ the claim follows easily.

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  • $\begingroup$ Can you please clarify why $|Im(z_k)/Re(z_k)|\le \tan \delta$? $\endgroup$ – sequence Feb 17 '17 at 4:13
  • $\begingroup$ @sequence Because $Im(z)/Re(z) = \tan \arg z$. This is nothing more than the formula that we learned in high school: The tangent equals opposite over adjacent (soh-cah-toa) $\endgroup$ – Shalop Feb 17 '17 at 4:17
  • $\begingroup$ It should be true that $|Im(z_k)/Re(z_k)| \le \delta$, correct? For if $z=x+iy$, then $\arg(z_k) = tan(y/x)\le tan\delta$, so $|y/x| \le \delta$. @Shalop $\endgroup$ – sequence Feb 17 '17 at 4:21
  • $\begingroup$ @sequence $\text{Re}(z)=|z|\cos(\arg(z))$ and $\text{Ie}(z)=|z|\sin(\arg(z))$. Hence, $\frac{\text{Im}(z)}{\text{Re}(z)}=\tan(\arg(z))$ $\endgroup$ – Mark Viola Feb 17 '17 at 4:24
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    $\begingroup$ @sequence No that's not correct because it is not true that $y/x <\delta$. $\endgroup$ – Shalop Feb 17 '17 at 7:16

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