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Dr. CAN学习笔记-数学基础Ch0-7欧拉公式的证明
e
i
θ
=
cos
θ
+
sin
θ
i
,
i
=
−
1
e^{i\theta}=\cos \theta +\sin \theta i,i=\sqrt{-1}
eiθ=cosθ+sinθi,i=−1
证明:
f
(
θ
)
=
e
i
θ
cos
θ
+
sin
θ
i
f
′
(
θ
)
=
i
e
i
θ
(
cos
θ
+
sin
θ
i
)
−
e
i
θ
(
−
sin
θ
+
cos
θ
i
)
(
cos
θ
+
sin
θ
i
)
2
=
0
⇒
f
(
θ
)
=
c
o
n
s
tan
t
f
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θ
)
=
f
(
0
)
=
e
i
0
cos
0
+
sin
0
i
=
1
⇒
e
i
θ
cos
θ
+
sin
θ
i
=
1
⇒
e
i
θ
=
cos
θ
+
sin
θ
i
f\left( \theta \right) =\frac{e^{i\theta}}{\cos \theta +\sin \theta i} \\ f^{\prime}\left( \theta \right) =\frac{ie^{i\theta}\left( \cos \theta +\sin \theta i \right) -e^{i\theta}\left( -\sin \theta +\cos \theta i \right)}{\left( \cos \theta +\sin \theta i \right) ^2}=0 \\ \Rightarrow f\left( \theta \right) =\mathrm{cons}\tan\mathrm{t} \\ f\left( \theta \right) =f\left( 0 \right) =\frac{e^{i0}}{\cos 0+\sin 0i}=1\Rightarrow \frac{e^{i\theta}}{\cos \theta +\sin \theta i}=1 \\ \Rightarrow e^{i\theta}=\cos \theta +\sin \theta i
f(θ)=cosθ+sinθieiθf′(θ)=(cosθ+sinθi)2ieiθ(cosθ+sinθi)−eiθ(−sinθ+cosθi)=0⇒f(θ)=constantf(θ)=f(0)=cos0+sin0iei0=1⇒cosθ+sinθieiθ=1⇒eiθ=cosθ+sinθi
求解:
sin
x
=
2
\sin x=2
sinx=2 令:
sin
z
=
2
=
c
,
z
∈
C
\sin z=2=c,z\in \mathbb{C}
sinz=2=c,z∈C
{
e
i
z
=
cos
z
+
sin
z
i
e
i
(
−
z
)
=
cos
z
−
sin
z
i
⇒
e
i
z
−
e
−
i
z
=
2
sin
z
i
\begin{cases} e^{iz}=\cos z+\sin zi\\ e^{i\left( -z \right)}=\cos z-\sin zi\\ \end{cases}\Rightarrow e^{iz}-e^{-iz}=2\sin zi
{eiz=cosz+sinziei(−z)=cosz−sinzi⇒eiz−e−iz=2sinzi
∴
sin
z
=
e
i
z
−
e
−
i
z
2
i
=
c
⇒
e
a
i
−
b
−
e
b
−
a
i
2
i
=
e
a
i
e
−
b
−
e
b
e
−
a
i
2
i
=
c
\therefore \sin z=\frac{e^{iz}-e^{-iz}}{2i}=c\Rightarrow \frac{e^{ai-b}-e^{b-ai}}{2i}=\frac{e^{ai}e^{-b}-e^be^{-ai}}{2i}=c
∴sinz=2ieiz−e−iz=c⇒2ieai−b−eb−ai=2ieaie−b−ebe−ai=c 且有:
{
e
i
a
=
cos
a
+
sin
a
i
e
i
(
−
a
)
=
cos
a
−
sin
a
i
\begin{cases} e^{ia}=\cos a+\sin ai\\ e^{i\left( -a \right)}=\cos a-\sin ai\\ \end{cases}
{eia=cosa+sinaiei(−a)=cosa−sinai
⇒
e
−
b
(
cos
a
+
sin
a
i
)
−
e
b
(
cos
a
−
sin
a
i
)
2
i
=
(
e
−
b
−
e
b
)
cos
a
−
(
e
−
b
+
e
b
)
sin
a
i
2
i
=
c
⇒
1
2
(
e
b
−
e
−
b
)
cos
a
i
+
1
2
(
e
−
b
+
e
b
)
sin
a
=
c
=
c
+
0
i
\Rightarrow \frac{e^{-b}\left( \cos a+\sin ai \right) -e^b\left( \cos a-\sin ai \right)}{2i}=\frac{\left( e^{-b}-e^b \right) \cos a-\left( e^{-b}+e^b \right) \sin ai}{2i}=c \\ \Rightarrow \frac{1}{2}\left( e^b-e^{-b} \right) \cos ai+\frac{1}{2}\left( e^{-b}+e^b \right) \sin a=c=c+0i
⇒2ie−b(cosa+sinai)−eb(cosa−sinai)=2i(e−b−eb)cosa−(e−b+eb)sinai=c⇒21(eb−e−b)cosai+21(e−b+eb)sina=c=c+0i
⇒
{
1
2
(
e
−
b
+
e
b
)
sin
a
=
c
1
2
(
e
b
−
e
−
b
)
cos
a
=
0
\Rightarrow \begin{cases} \frac{1}{2}\left( e^{-b}+e^b \right) \sin a=c\\ \frac{1}{2}\left( e^b-e^{-b} \right) \cos a=0\\ \end{cases}
⇒{21(e−b+eb)sina=c21(eb−e−b)cosa=0
当
b
=
0
b=0
b=0 时,
sin
a
=
c
\sin a=c
sina=c 不成立(所设
a
,
b
∈
R
a,b\in \mathbb{R}
a,b∈R)当
cos
a
=
0
\cos a=0
cosa=0 时,
1
2
(
e
−
b
+
e
b
)
=
±
c
⇒
1
+
e
2
b
±
2
c
e
b
=
0
\frac{1}{2}\left( e^{-b}+e^b \right) =\pm c\Rightarrow 1+e^{2b}\pm 2ce^b=0
21(e−b+eb)=±c⇒1+e2b±2ceb=0 设
u
=
e
b
>
0
u=e^b>0
u=eb>0 ,则有:
u
=
±
c
±
c
2
−
1
u=\pm c\pm \sqrt{c^2-1}
u=±c±c2−1
∴
b
=
ln
(
c
±
c
2
−
1
)
\therefore b=\ln \left( c\pm \sqrt{c^2-1} \right)
∴b=ln(c±c2−1
)
⇒
z
=
π
2
+
2
k
π
+
ln
(
c
±
c
2
−
1
)
i
=
π
2
+
2
k
π
+
ln
(
2
±
3
)
i
\Rightarrow z=\frac{\pi}{2}+2k\pi +\ln \left( c\pm \sqrt{c^2-1} \right) i=\frac{\pi}{2}+2k\pi +\ln \left( 2\pm \sqrt{3} \right) i
⇒z=2π+2kπ+ln(c±c2−1
)i=2π+2kπ+ln(2±3
)i
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