Mixing
Prove that $sin(-x)=-sin(x)$ and $cos(-x)=cos(x)$ using complex conjugation
$begingroup$
The Definition of $sin$ and $cos$ is given by the complex function,
$$f:mathbb{C}rightarrowmathbb{C},qquad
f(x)=e^{text{i}x}=cos(x)+text{i}sin(x)$$
I.e. cos is the real part of $e^{ix}$ and sin denotes the immaginary part of $e^{ix}$.
Now we know that $forall z_iinmathbb{C}$
$$overline{sum_{iin I}z_i}=sum_{iin I}bar{z_i}$$
therefore since $e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$
$$overline{e^{ix}}=e^{overline{ix}}$$
for the left side I would get
$cos(x)-text{i}sin(x)$
Can someone explain why we have for the right side
$cos(-x)+ text{i}sin(x)$
And also how do we derive from the Formula
$$cos(-x)+ text{i}sin(x)=cos(x)-text{i}sin(x)$$
that we got from above that
$sin(-x)=sin(x)$
The equation
$cos(-x)=cos(x)$
is clear
complex-analysis
edited Jan 22 at 21:35
egreg
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
The Definition of $sin$ and $cos$ is given by the complex function,
$$f:mathbb{C}rightarrowmathbb{C},qquad
f(x)=e^{text{i}x}=cos(x)+text{i}sin(x)$$
I.e. cos is the real part of $e^{ix}$ and sin denotes the immaginary part of $e^{ix}$.
Now we know that $forall z_iinmathbb{C}$
$$overline{sum_{iin I}z_i}=sum_{iin I}bar{z_i}$$
therefore since $e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$
$$overline{e^{ix}}=e^{overline{ix}}$$
for the left side I would get
$cos(x)-text{i}sin(x)$
Can someone explain why we have for the right side
$cos(-x)+ text{i}sin(x)$
And also how do we derive from the Formula
$$cos(-x)+ text{i}sin(x)=cos(x)-text{i}sin(x)$$
that we got from above that
$sin(-x)=sin(x)$
The equation
$cos(-x)=cos(x)$
is clear
complex-analysis
edited Jan 22 at 21:35
egreg
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
$endgroup$
1
$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38
add a comment |
$begingroup$
The Definition of $sin$ and $cos$ is given by the complex function,
$$f:mathbb{C}rightarrowmathbb{C},qquad
f(x)=e^{text{i}x}=cos(x)+text{i}sin(x)$$
I.e. cos is the real part of $e^{ix}$ and sin denotes the immaginary part of $e^{ix}$.
Now we know that $forall z_iinmathbb{C}$
$$overline{sum_{iin I}z_i}=sum_{iin I}bar{z_i}$$
therefore since $e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$
$$overline{e^{ix}}=e^{overline{ix}}$$
for the left side I would get
$cos(x)-text{i}sin(x)$
Can someone explain why we have for the right side
$cos(-x)+ text{i}sin(x)$
And also how do we derive from the Formula
$$cos(-x)+ text{i}sin(x)=cos(x)-text{i}sin(x)$$
that we got from above that
$sin(-x)=sin(x)$
The equation
$cos(-x)=cos(x)$
is clear
complex-analysis
edited Jan 22 at 21:35
egreg
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
$endgroup$
The Definition of $sin$ and $cos$ is given by the complex function,
$$f:mathbb{C}rightarrowmathbb{C},qquad
f(x)=e^{text{i}x}=cos(x)+text{i}sin(x)$$
I.e. cos is the real part of $e^{ix}$ and sin denotes the immaginary part of $e^{ix}$.
Now we know that $forall z_iinmathbb{C}$
$$overline{sum_{iin I}z_i}=sum_{iin I}bar{z_i}$$
therefore since $e^x:=sum_{k=0}^{infty}frac{x^k}{k!}$
$$overline{e^{ix}}=e^{overline{ix}}$$
for the left side I would get
$cos(x)-text{i}sin(x)$
Can someone explain why we have for the right side
$cos(-x)+ text{i}sin(x)$
And also how do we derive from the Formula
$$cos(-x)+ text{i}sin(x)=cos(x)-text{i}sin(x)$$
that we got from above that
$sin(-x)=sin(x)$
The equation
$cos(-x)=cos(x)$
is clear
complex-analysis
complex-analysis
edited Jan 22 at 21:35
egreg
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
edited Jan 22 at 21:35
egreg
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
edited Jan 22 at 21:35
egreg
184k1486205
edited Jan 22 at 21:35
egreg
184k1486205
edited Jan 22 at 21:35
egreg
184k1486205
184k1486205
asked Jan 22 at 21:25
RM777RM777
38312
asked Jan 22 at 21:25
RM777RM777
38312
asked Jan 22 at 21:25
RM777RM777
38312
38312
1
$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38
add a comment |
1
$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38
1
1
$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not true that $sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
overline{ix}=bar{i}bar{x}=-ix
$$
Therefore, after noting that $overline{e^z}=e^{bar{z}}$ as you did,
$$
overline{cos x+isin x}=overline{e^{ix}}=e^{overline{ix}}=e^{-ix}=cos(-x)+isin(-x)
$$
Hence
$$
cos x-isin x=cos(-x)+isin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
overline{cos x+isin x}necos x-isin x
$$
in general.
On the other hand, the definition of $sin z$ for complex $z$ is
$$
sin z=frac{e^{iz}-e^{-iz}}{2i}
$$
and $sin(-z)=-sin z$ is obvious.
For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(cos a+isin a)$, so the real part of $e^{iz}$ is $e^{-b}cos a$, which is not $cos z$ in general, for the simple reason that $cos z$ is not itself real in general and you can observe that the real part is $cos z$ if and only if $b=0$, that is, if $z$ is real.
answered Jan 22 at 21:39
egregegreg
184k1486205
$endgroup$
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
add a comment |
$begingroup$
Usually that first equation is a theorem (Euler's formula) rather than a definition. But if you start from there, notice:
$$
e^{-ix} = e^{i(-x)} = cos(-x) + i sin(-x)
$$
From the power series formula $e^z = sum_{k=0}^infty frac{z^k}{k!}$, we see that
$$
e^{-ix} = sum_{k=0}^infty frac{(-ix)^k}{k!} = sum_{k=0}^infty frac{(overline{ix})^k}{k!} = overline{e^{ix}}
$$
Therefore
$$
e^{-ix} = overline{e^{ix}} = overline{cos(x) + i sin(x)} = cos(x) - i sin(x)
$$
Now you just equate real and imaginary parts.
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not true that $sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
overline{ix}=bar{i}bar{x}=-ix
$$
Therefore, after noting that $overline{e^z}=e^{bar{z}}$ as you did,
$$
overline{cos x+isin x}=overline{e^{ix}}=e^{overline{ix}}=e^{-ix}=cos(-x)+isin(-x)
$$
Hence
$$
cos x-isin x=cos(-x)+isin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
overline{cos x+isin x}necos x-isin x
$$
in general.
On the other hand, the definition of $sin z$ for complex $z$ is
$$
sin z=frac{e^{iz}-e^{-iz}}{2i}
$$
and $sin(-z)=-sin z$ is obvious.
For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(cos a+isin a)$, so the real part of $e^{iz}$ is $e^{-b}cos a$, which is not $cos z$ in general, for the simple reason that $cos z$ is not itself real in general and you can observe that the real part is $cos z$ if and only if $b=0$, that is, if $z$ is real.
answered Jan 22 at 21:39
egregegreg
184k1486205
$endgroup$
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
add a comment |
$begingroup$
It is not true that $sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
overline{ix}=bar{i}bar{x}=-ix
$$
Therefore, after noting that $overline{e^z}=e^{bar{z}}$ as you did,
$$
overline{cos x+isin x}=overline{e^{ix}}=e^{overline{ix}}=e^{-ix}=cos(-x)+isin(-x)
$$
Hence
$$
cos x-isin x=cos(-x)+isin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
overline{cos x+isin x}necos x-isin x
$$
in general.
On the other hand, the definition of $sin z$ for complex $z$ is
$$
sin z=frac{e^{iz}-e^{-iz}}{2i}
$$
and $sin(-z)=-sin z$ is obvious.
For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(cos a+isin a)$, so the real part of $e^{iz}$ is $e^{-b}cos a$, which is not $cos z$ in general, for the simple reason that $cos z$ is not itself real in general and you can observe that the real part is $cos z$ if and only if $b=0$, that is, if $z$ is real.
answered Jan 22 at 21:39
egregegreg
184k1486205
$endgroup$
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
add a comment |
$begingroup$
It is not true that $sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
overline{ix}=bar{i}bar{x}=-ix
$$
Therefore, after noting that $overline{e^z}=e^{bar{z}}$ as you did,
$$
overline{cos x+isin x}=overline{e^{ix}}=e^{overline{ix}}=e^{-ix}=cos(-x)+isin(-x)
$$
Hence
$$
cos x-isin x=cos(-x)+isin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
overline{cos x+isin x}necos x-isin x
$$
in general.
On the other hand, the definition of $sin z$ for complex $z$ is
$$
sin z=frac{e^{iz}-e^{-iz}}{2i}
$$
and $sin(-z)=-sin z$ is obvious.
For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(cos a+isin a)$, so the real part of $e^{iz}$ is $e^{-b}cos a$, which is not $cos z$ in general, for the simple reason that $cos z$ is not itself real in general and you can observe that the real part is $cos z$ if and only if $b=0$, that is, if $z$ is real.
answered Jan 22 at 21:39
egregegreg
184k1486205
$endgroup$
It is not true that $sin x$ is the imaginary part of $e^{ix}$, if $x$ is complex nonreal.
If $x$ is real, then
$$
overline{ix}=bar{i}bar{x}=-ix
$$
Therefore, after noting that $overline{e^z}=e^{bar{z}}$ as you did,
$$
overline{cos x+isin x}=overline{e^{ix}}=e^{overline{ix}}=e^{-ix}=cos(-x)+isin(-x)
$$
Hence
$$
cos x-isin x=cos(-x)+isin(-x)
$$
and the equality is proved.
Note that when $x$ is complex the derivation is slightly different, mainly because
$$
overline{cos x+isin x}necos x-isin x
$$
in general.
On the other hand, the definition of $sin z$ for complex $z$ is
$$
sin z=frac{e^{iz}-e^{-iz}}{2i}
$$
and $sin(-z)=-sin z$ is obvious.
For completeness, if $z=a+ib$, then $e^{iz}=e^{-b+ia}=e^{-b}(cos a+isin a)$, so the real part of $e^{iz}$ is $e^{-b}cos a$, which is not $cos z$ in general, for the simple reason that $cos z$ is not itself real in general and you can observe that the real part is $cos z$ if and only if $b=0$, that is, if $z$ is real.
answered Jan 22 at 21:39
egregegreg
184k1486205
edited Jan 22 at 21:49
answered Jan 22 at 21:39
egregegreg
184k1486205
answered Jan 22 at 21:39
egregegreg
184k1486205
answered Jan 22 at 21:39
egregegreg
184k1486205
184k1486205
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
add a comment |
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
$begingroup$
if $bar{i}= -i$, I only get $bar{i}bar{x}=-ibar{x}$, why does the bar of x vanish too?
$endgroup$
– RM777
Jan 22 at 21:42
1
1
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
@RM777 For complex $x$, it is not true that $sin x$ is the imaginary part of $e^{Ix}$. That derivation is only valid for real $x$.
$endgroup$
– egreg
Jan 22 at 21:43
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
$begingroup$
How do I know if $e^{text{i}x}=cos(x)+text{i}sin(x)$ that $e^{text{-i}x}=cos(-x)+text{i}sin(-x)$?
$endgroup$
– RM777
Jan 22 at 21:59
1
1
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
$begingroup$
@RM777 Change $x$ into $-x$.
$endgroup$
– egreg
Jan 22 at 22:04
add a comment |
$begingroup$
Usually that first equation is a theorem (Euler's formula) rather than a definition. But if you start from there, notice:
$$
e^{-ix} = e^{i(-x)} = cos(-x) + i sin(-x)
$$
From the power series formula $e^z = sum_{k=0}^infty frac{z^k}{k!}$, we see that
$$
e^{-ix} = sum_{k=0}^infty frac{(-ix)^k}{k!} = sum_{k=0}^infty frac{(overline{ix})^k}{k!} = overline{e^{ix}}
$$
Therefore
$$
e^{-ix} = overline{e^{ix}} = overline{cos(x) + i sin(x)} = cos(x) - i sin(x)
$$
Now you just equate real and imaginary parts.
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
add a comment |
$begingroup$
Usually that first equation is a theorem (Euler's formula) rather than a definition. But if you start from there, notice:
$$
e^{-ix} = e^{i(-x)} = cos(-x) + i sin(-x)
$$
From the power series formula $e^z = sum_{k=0}^infty frac{z^k}{k!}$, we see that
$$
e^{-ix} = sum_{k=0}^infty frac{(-ix)^k}{k!} = sum_{k=0}^infty frac{(overline{ix})^k}{k!} = overline{e^{ix}}
$$
Therefore
$$
e^{-ix} = overline{e^{ix}} = overline{cos(x) + i sin(x)} = cos(x) - i sin(x)
$$
Now you just equate real and imaginary parts.
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
add a comment |
$begingroup$
Usually that first equation is a theorem (Euler's formula) rather than a definition. But if you start from there, notice:
$$
e^{-ix} = e^{i(-x)} = cos(-x) + i sin(-x)
$$
From the power series formula $e^z = sum_{k=0}^infty frac{z^k}{k!}$, we see that
$$
e^{-ix} = sum_{k=0}^infty frac{(-ix)^k}{k!} = sum_{k=0}^infty frac{(overline{ix})^k}{k!} = overline{e^{ix}}
$$
Therefore
$$
e^{-ix} = overline{e^{ix}} = overline{cos(x) + i sin(x)} = cos(x) - i sin(x)
$$
Now you just equate real and imaginary parts.
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
$endgroup$
Usually that first equation is a theorem (Euler's formula) rather than a definition. But if you start from there, notice:
$$
e^{-ix} = e^{i(-x)} = cos(-x) + i sin(-x)
$$
From the power series formula $e^z = sum_{k=0}^infty frac{z^k}{k!}$, we see that
$$
e^{-ix} = sum_{k=0}^infty frac{(-ix)^k}{k!} = sum_{k=0}^infty frac{(overline{ix})^k}{k!} = overline{e^{ix}}
$$
Therefore
$$
e^{-ix} = overline{e^{ix}} = overline{cos(x) + i sin(x)} = cos(x) - i sin(x)
$$
Now you just equate real and imaginary parts.
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
edited Jan 22 at 21:52
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
answered Jan 22 at 21:31
Matthew LeingangMatthew Leingang
16.8k12244
16.8k12244
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
add a comment |
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
$begingroup$
Why is $e^{-ix}=overline{e^{ix}}$?
$endgroup$
– RM777
Jan 22 at 21:34
1
1
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
$begingroup$
$overline i = -i$
$endgroup$
– J. W. Tanner
Jan 22 at 21:38
1
1
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
@RM777 I've included a line of justification for that formula.
$endgroup$
– Matthew Leingang
Jan 22 at 21:53
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
$begingroup$
Last sentence was especially helpful I was still wondering how I get the Formula untill now. Thanks
$endgroup$
– RM777
Jan 22 at 22:05
add a comment |
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$begingroup$
It's $overline{e^{ix}}=e^{-ix}=e^{i(-x)}$.
$endgroup$
– Isko10986
Jan 22 at 21:30
$begingroup$
@Isko10986 why is $e^{-ix}=overline{e^{ix}}$
$endgroup$
– RM777
Jan 22 at 21:38