4.3 參考解答

  1. Σn=0∞(z+i)n2n{converges D2(−i)={z:∣z+i∣<2}diverges ∣z+i∣>2\Sigma_{n=0}^{\infty} \frac{(z+i)^n}{2^n} \begin{cases} \text{converges}\space D_2(-i)=\{z: |z+i|<2 \} \\ \text{diverges}\space |z+i|>2\end{cases}

    lim⁡n→∞∣(z+i)n+12n+1×2n(z+i)n∣=lim⁡n→∞∣z+i2∣=∣z+i2∣\lim_{n \to \infty} |\frac{(z+i)^{n+1}}{2^{n+1}} \times \frac{2^n}{(z+i)^n}|=\lim_{n \to \infty} |\frac{z+i}{2}|=|\frac{z+i}{2}|

    if ∣z+i2∣<1⇒∣z+i∣<2⇒Σn=0∞(z+i)n2n|\frac{z+i}{2}|<1 \Rightarrow |z+i|<2 \Rightarrow \Sigma_{n=0}^{\infty} \frac{(z+i)^n}{2^n} is convergent. (by Ratio Test)

    if ∣z+i2∣>1⇒∣z+i∣>2⇒Σn=0∞(z+i)n2n|\frac{z+i}{2}|>1 \Rightarrow |z+i|>2 \Rightarrow \Sigma_{n=0}^{\infty} \frac{(z+i)^n}{2^n} is divergent.

4. (a) Σn=0∞(1+i2)n\Sigma_{n=0}^{\infty} (\frac{1+i}{2})^n is convergent (by Ratio Test)

lim⁡n→∞∣(1+i2)n+1×(21+i)n∣=lim⁡n→∞∣1+i2∣=22<1\lim_{n \to \infty} |(\frac{1+i}{2})^{n+1} \times (\frac{2}{1+i})^n|=\lim_{n \to \infty} |\frac{1+i}{2}|=\frac{\sqrt{2}}{2}<1

(c) Σn=1∞(1+i)nn!\Sigma_{n=1}^{\infty} \frac{(1+i)^n}{n!} is convergent (by Ratio Test)

lim⁡n→∞∣(1+i)n+1(n+1)!×n!(1+i)n∣=lim⁡n→∞∣1+in+1∣=lim⁡n→∞2n+1=0<1\lim_{n \to \infty} |\frac{(1+i)^{n+1}}{(n+1)!} \times \frac{n!}{(1+i)^n}|=\lim_{n \to \infty} |\frac{1+i}{n+1}|=\lim_{n \to \infty} \frac{\sqrt{2}}{n+1}=0<1

  1. (a) Σn=0∞(1+i)nzn\Sigma_{n=0}^{\infty} (1+i)^nz^n is convergent on D12(0)D_{\frac{1}{\sqrt{2}}}(0)

    lim⁡n→∞∣zn+1(1+i)n+1zn(1+i)n∣=lim⁡n→∞∣z(1+i)∣=2∣z∣\lim_{n \to \infty} |\frac{z^{n+1}(1+i)^{n+1}}{z^n(1+i)^n}|=\lim_{n \to \infty} |z(1+i)|=\sqrt{2}|z|

    2∣z∣<1⇒∣z∣<12\sqrt{2}|z|<1 \Rightarrow |z|<\frac{1}{\sqrt{2}}

    (c) Σn=0∞(z−i)n(3+4i)n\Sigma_{n=0}^{\infty} \frac{(z-i)^n}{(3+4i)^n} is convergent D5(i)D_5(i)

    lim⁡n→∞∣(z−i)n+1(3+4i)n+1×(3+4i)n(z−i)n∣=lim⁡n→∞∣z−i3+4i∣=∣z−i∣5\lim_{n \to \infty} |\frac{(z-i)^{n+1}}{(3+4i)^{n+1}} \times \frac{(3+4i)^n}{(z-i)^n}|=\lim_{n \to \infty} |\frac{z-i}{3+4i}|=\frac{|z-i|}{5}

    ∣z−i∣5<1⇒∣z−i∣<5\frac{|z-i|}{5}<1 \Rightarrow |z-i|<5

12. (a) z=reiθ z=re^{i\theta}, ∣z∣=r<1|z|=r<1

By geometric series, Σn=0∞zn=11−z\Sigma_{n=0}^{\infty} z^n=\frac{1}{1-z}

Σn=0∞zn=Σn=0∞(reiθ)n=Σn=0∞rneinθ\Sigma_{n=0}^{\infty} z^n=\Sigma_{n=0}^{\infty} (re^{i\theta})^n=\Sigma_{n=0}^{\infty} r^n e^{in\theta}

=11−(rcos⁡θ+irsin⁡θ)×(1−rcos⁡θ)+irsin⁡θ(1−rcos⁡θ)+irsin⁡θ=\frac{1}{1-(r\cos{\theta}+ir\sin{\theta})} \times \frac{(1-r\cos{\theta})+ir\sin{\theta}}{(1-r\cos{\theta})+ir\sin{\theta}}

=1−rcos⁡θ+irsin⁡θ1+r2−2rcos⁡θ=\frac{1-r\cos{\theta}+ir\sin{\theta}}{1+r^2-2r\cos{\theta}}

(b) ∵Σn=0∞zn=Σn=0∞(reiθ)n=Σn=0∞rneinθ\because \Sigma_{n=0}^{\infty} z^n=\Sigma_{n=0}^{\infty} (re^{i\theta})^n=\Sigma_{n=0}^{\infty} r^n e^{in\theta}

=Σn=0∞rn(cos⁡nθ+isin⁡nθ)=Σn=0∞(rncos⁡nθ+irnsin⁡nθ) =\Sigma_{n=0}^{\infty} r^n(\cos{n\theta}+i\sin{n\theta})=\Sigma_{n=0}^{\infty} (r^n\cos{n\theta}+ir^n\sin{n\theta})

∴1−rcos⁡θ+irsin⁡θ1+r2−2rcos⁡θ=1−rcos⁡θ1+r2−2rcos⁡θ+irsin⁡θ1+r2−2rcos⁡θ\therefore \frac{1-r\cos{\theta}+ir\sin{\theta}}{1+r^2-2r\cos{\theta}}=\frac{1-r\cos{\theta}}{1+r^2-2r\cos{\theta}}+i\frac{r\sin{\theta}}{1+r^2-2r\cos{\theta}}

Thus, Σn=0∞rncos⁡nθ=1−rcos⁡θ1+r2−2rcos⁡θ\Sigma_{n=0}^{\infty} r^n\cos{n\theta}=\frac{1-r\cos{\theta}}{1+r^2-2r\cos{\theta}}  and Σn=0∞rnsin⁡nθ=rsin⁡θ1+r2−2rcos⁡θ\Sigma_{n=0}^{\infty} r^n\sin{n\theta}=\frac{r\sin{\theta}}{1+r^2-2r\cos{\theta}}.