1.4 參考解答

  1. (a) −4=4eiπ-4=4e^{i\pi}

    (b) 6−6i=62ei(−π4)6-6i=6\sqrt{2}e^{i(-\frac{\pi}{4})}

    (c) −7i=7ei(−π2)-7i=7e^{i(-\frac{\pi}{2})}

    (d) −23−2i=4ei(−56π)-2\sqrt{3}-2i=4e^{i(-\frac{5}{6}\pi)}

    (e) 1(1−i)2=1(2ei(−π4))2=12ei(−π2)=12eiπ2\frac{1}{(1-i)^2}=\frac{1}{(\sqrt{2}e^{i(-\frac{\pi}{4})})^2}=\frac{1}{2e^{i(-\frac{\pi}{2})}}=\frac{1}{2}e^{i\frac{\pi}{2}}

    (f) 6i+3=6ei⋅02eiπ6=3ei(−π6)\frac{6}{i+\sqrt{3}}=\frac{6e^{i \cdot 0}}{2e^{i\frac{\pi}{6}}}=3e^{i(-\frac{\pi}{6})}

    (g) 3+4i=5eiθ,θ=tan−1 (43)3+4i=5e^{i\theta}, \theta=\text{tan}^{-1}\space (\frac{4}{3})

    (h) (5+5i)3=(52eiπ4)3=2502ei3π4(5+5i)^3=(5\sqrt{2}e^{i\frac{\pi}{4}})^3=250\sqrt{2}e^{i\frac{3\pi}{4}}

5. (a) eiπ2=cos⁡π2+isin⁡π2=ie^{i\frac{\pi}{2}}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i

(b) 4e−iπ2=4(cos⁡(−π2)+isin⁡(−π2))=4(cos⁡π2−isin⁡π2)=4(−i)=−4i4e^{-i\frac{\pi}{2}}=4(\cos{(-\frac{\pi}{2})}+i\sin{(-\frac{\pi}{2})})=4(\cos{\frac{\pi}{2}}-i\sin{\frac{\pi}{2}})=4(-i)=-4i

(c) 8ei7π3=8(cos⁡7π3+isin⁡7π3)=8(cos⁡π3+isin⁡π3)=8(12+i32)=4+i438e^{i\frac{7\pi}{3}}=8(\cos{\frac{7\pi}{3}}+i\sin{\frac{7\pi}{3}})=8(\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}})=8(\frac{1}{2}+i\frac{\sqrt{3}}{2})=4+i4\sqrt{3}

(d) −2ei5π6=−2(cos⁡5π6+isin⁡5π6)=−2(−32+i12)=3−i-2e^{i\frac{5\pi}{6}}=-2(\cos{\frac{5\pi}{6}}+i\sin {\frac{5\pi}{6}})=-2(-\frac{\sqrt{3}}{2}+i\frac{1}{2})=\sqrt{3}-i

(e) 2ie−i3π4=2i(cos⁡(−3π4)+isin⁡(−3π4))=2i(cos⁡3π4−isin⁡3π4)2ie^{-i\frac{3\pi}{4}}=2i(\cos{(-\frac{3\pi}{4}})+i\sin{(-\frac{3\pi}{4})})=2i(\cos{\frac{3\pi}{4}}-i\sin{\frac{3\pi}{4}})

=2i(−22−i22)=2−i2=2i(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=\sqrt{2}-i\sqrt{2}

(f) 6ei2π3eiπ=6ei5π3=6(cos⁡5π3+isin⁡5π3)=6(cos⁡(−π3)+isin⁡(−π3))6e^{i\frac{2\pi}{3}}e^{i\pi}=6e^{i\frac{5\pi}{3}}=6(\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}})=6(\cos{(-\frac{\pi}{3}})+i\sin{(-\frac{\pi}{3})})

=6(cos⁡π3−isin⁡π3)=6(12−i32)=3−i33=6(\cos{\frac{\pi}{3}}-i\sin{\frac{\pi}{3}})=6(\frac{1}{2}-i\frac{\sqrt{3}}{2})=3-i3\sqrt{3}

(g) e2eiπ=e2(cos⁡π+isin⁡π)=e2(−1)=−e2e^2e^{i\pi}=e^2(\cos{\pi}+i\sin{\pi})=e^2(-1)=-e^2

(h) eiπ4e−iπ=e−i3π4=cos⁡(−3π4)+isin⁡(−3π4)=cos⁡3π4−isin⁡3π4=−22−i22e^{i\frac{\pi}{4}}e^{-i\pi}=e^{-i\frac{3\pi}{4}}=\cos{(-\frac{3\pi}{4})}+i\sin{(-\frac{3\pi}{4})}=\cos{\frac{3\pi}{4}}-i\sin{\frac{3\pi}{4}}=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}

  1. “⇒\Rightarrow” If arg z1=arg z2\text{arg}\space z_1=\text{arg}\space z_2 for z1≠z2z_1 \neq z_2,

then θ∈arg⁡z1\theta \in \arg{z_1}, where z1=∣z1∣eiθz_1=|z_1|e^{i\theta}

θ∈arg⁡z2\theta \in \arg{z_2}, where z2=∣z2∣eiθz_2=|z_2|e^{i\theta}

z2=∣z2∣eiθ=∣z2∣⋅∣z1∣∣z1∣eiθ=∣z2z1∣⋅∣z1∣eiθ=c⋅z1z_2=|z_2|e^{i\theta}=|z_2|\cdot\frac{|z_1|}{|z_1|}e^{i\theta}=|\frac{z_2}{z_1}|\cdot|z_1|e^{i\theta}=c\cdot z_1, where c=∣z2z1∣>0c=|\frac{z_2}{z_1}|>0.

“⇐\Leftarrow” If z2=cz1z_2=cz_1, c>0c>0, then we have

∣z2∣=∣cz1∣=c∣z1∣|z_2|=|cz_1|=c|z_1|, ∣z1∣=1c∣z2∣|z_1|=\frac{1}{c}|z_2|

1。1^。 If θ∈arg⁡z1\theta \in \arg{z_1}, where z1=∣z1∣eiθz_1=|z_1|e^{i\theta},

then z2=cz1=c∣z1∣eiθ=∣z2∣eiθz_2=cz_1=c|z_1|e^{i\theta}=|z_2|e^{i\theta},

so θ∈arg⁡z2\theta \in \arg{z_2}, then arg⁡z1⊆arg⁡z2\arg{z_1} \subseteq \arg{z_2}.

2。2^。 If θ∈arg⁡z2\theta \in \arg{z_2}, where z2=∣z2∣eiθz_2=|z_2|e^{i\theta},

then z1=1cz2=1c∣z2∣eiθ=∣z1∣eiθz_1=\frac{1}{c} z_2=\frac{1}{c} |z_2|e^{i\theta}=|z_1|e^{i\theta},

so θ∈arg⁡z1\theta \in \arg{z_1}, then arg⁡z2⊆arg⁡z1\arg{z_2} \subseteq \arg{z_1}.

  1. counterexample:

    Remark: arg⁡(z1)+arg⁡(z2)=arg⁡(z1z2)\arg{(z_1)}+\arg{(z_2)}=\arg{(z_1z_2)}

    Arg (z1)+Arg (z2)≠Arg (z1z2)\text{Arg}\space (z_1)+\text{Arg}\space (z_2) \neq \text{Arg}\space (z_1z_2)

    z1=−1+i3z_1=-1+i\sqrt{3}, Arg (z1)=2π3\text{Arg}\space (z_1)=\frac{2\pi}{3}

    z1=−3+iz_1=-\sqrt{3}+i, Arg (z2)=5π6\text{Arg}\space (z_2)=\frac{5\pi}{6}

    z1⋅z2=2ei2π3⋅2ei5π6=4ei3π2z_1\cdot z_2=2e^{i\frac{2\pi}{3}} \cdot 2e^{i\frac{5\pi}{6}}=4e^{i\frac{3\pi}{2}}

    And Arg (z1z2)=−π2\text{Arg}\space (z_1z_2)=-\frac{\pi}{2}

    Thus, Arg (z1z2)≠Arg (z1)+Arg (z2)\text{Arg}\space (z_1z_2) \neq \text{Arg}\space (z_1)+\text{Arg}\space (z_2)