Complex exponential function
$begingroup$
Just a basic question. I have $e^{-i9pi/4}$, and I'm struggling to understand why this is equal to $e^{-ipi/4}$.
Thanks in advance for any help.
complex-numbers
edited May 23 '11 at 17:38
Namaste
1
asked May 23 '11 at 14:07
DooodleDooodle
312
$endgroup$
add a comment |
$begingroup$
Just a basic question. I have $e^{-i9pi/4}$, and I'm struggling to understand why this is equal to $e^{-ipi/4}$.
Thanks in advance for any help.
complex-numbers
edited May 23 '11 at 17:38
Namaste
1
asked May 23 '11 at 14:07
DooodleDooodle
312
$endgroup$
4
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10
add a comment |
$begingroup$
Just a basic question. I have $e^{-i9pi/4}$, and I'm struggling to understand why this is equal to $e^{-ipi/4}$.
Thanks in advance for any help.
complex-numbers
edited May 23 '11 at 17:38
Namaste
1
asked May 23 '11 at 14:07
DooodleDooodle
312
$endgroup$
Just a basic question. I have $e^{-i9pi/4}$, and I'm struggling to understand why this is equal to $e^{-ipi/4}$.
Thanks in advance for any help.
complex-numbers
complex-numbers
edited May 23 '11 at 17:38
Namaste
1
asked May 23 '11 at 14:07
DooodleDooodle
312
edited May 23 '11 at 17:38
Namaste
1
asked May 23 '11 at 14:07
DooodleDooodle
312
edited May 23 '11 at 17:38
Namaste
1
edited May 23 '11 at 17:38
Namaste
1
edited May 23 '11 at 17:38
Namaste
1
1
asked May 23 '11 at 14:07
DooodleDooodle
312
asked May 23 '11 at 14:07
DooodleDooodle
312
asked May 23 '11 at 14:07
DooodleDooodle
312
312
4
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10
add a comment |
4
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10
4
4
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because
$e^{itheta}=costheta + icdot sintheta$,
$e^{-itheta}=cos(-theta)+icdot sin(-theta)=costheta - i sin(theta)$.
$sin(2pi+t)=sin(t)$, same for cosine also.
$sinleft(frac{9pi}{4}right)=sin left(2pi + frac{pi}{4}right)$.
So that
begin{align*}
e^{-ifrac{9pi}{4}}
&=cosfrac{-9pi}{4}+i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} +i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} -isinfrac{9pi}{4}\
&= cosleft(2pi + frac{pi}{4}right)-isinleft(2pi +frac{pi}{4}right)\
&=cosfrac{pi}{4}-isinfrac{pi}{4}\
&=e^{-ifrac{pi}{4}}
end{align*}
Also while doing such problems, these formulae may come handy:
$sin(pi+theta)= - sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sin(pi-theta) = sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2}+thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2} -thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi+theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi - theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2}-thetabigr)=sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2} + thetabigr)= -sintheta ; qquad 0 leq theta leq frac{pi}{2}$
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
$endgroup$
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
|
show 4 more comments
$begingroup$
Note that $9pi/4 = (1+8)left(pi/4right) = pi/4 + 2pi$ and $e^{2pi i}=1$.
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f40835%2fcomplex-exponential-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because
$e^{itheta}=costheta + icdot sintheta$,
$e^{-itheta}=cos(-theta)+icdot sin(-theta)=costheta - i sin(theta)$.
$sin(2pi+t)=sin(t)$, same for cosine also.
$sinleft(frac{9pi}{4}right)=sin left(2pi + frac{pi}{4}right)$.
So that
begin{align*}
e^{-ifrac{9pi}{4}}
&=cosfrac{-9pi}{4}+i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} +i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} -isinfrac{9pi}{4}\
&= cosleft(2pi + frac{pi}{4}right)-isinleft(2pi +frac{pi}{4}right)\
&=cosfrac{pi}{4}-isinfrac{pi}{4}\
&=e^{-ifrac{pi}{4}}
end{align*}
Also while doing such problems, these formulae may come handy:
$sin(pi+theta)= - sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sin(pi-theta) = sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2}+thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2} -thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi+theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi - theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2}-thetabigr)=sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2} + thetabigr)= -sintheta ; qquad 0 leq theta leq frac{pi}{2}$
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
$endgroup$
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
|
show 4 more comments
$begingroup$
Because
$e^{itheta}=costheta + icdot sintheta$,
$e^{-itheta}=cos(-theta)+icdot sin(-theta)=costheta - i sin(theta)$.
$sin(2pi+t)=sin(t)$, same for cosine also.
$sinleft(frac{9pi}{4}right)=sin left(2pi + frac{pi}{4}right)$.
So that
begin{align*}
e^{-ifrac{9pi}{4}}
&=cosfrac{-9pi}{4}+i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} +i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} -isinfrac{9pi}{4}\
&= cosleft(2pi + frac{pi}{4}right)-isinleft(2pi +frac{pi}{4}right)\
&=cosfrac{pi}{4}-isinfrac{pi}{4}\
&=e^{-ifrac{pi}{4}}
end{align*}
Also while doing such problems, these formulae may come handy:
$sin(pi+theta)= - sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sin(pi-theta) = sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2}+thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2} -thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi+theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi - theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2}-thetabigr)=sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2} + thetabigr)= -sintheta ; qquad 0 leq theta leq frac{pi}{2}$
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
$endgroup$
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
|
show 4 more comments
$begingroup$
Because
$e^{itheta}=costheta + icdot sintheta$,
$e^{-itheta}=cos(-theta)+icdot sin(-theta)=costheta - i sin(theta)$.
$sin(2pi+t)=sin(t)$, same for cosine also.
$sinleft(frac{9pi}{4}right)=sin left(2pi + frac{pi}{4}right)$.
So that
begin{align*}
e^{-ifrac{9pi}{4}}
&=cosfrac{-9pi}{4}+i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} +i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} -isinfrac{9pi}{4}\
&= cosleft(2pi + frac{pi}{4}right)-isinleft(2pi +frac{pi}{4}right)\
&=cosfrac{pi}{4}-isinfrac{pi}{4}\
&=e^{-ifrac{pi}{4}}
end{align*}
Also while doing such problems, these formulae may come handy:
$sin(pi+theta)= - sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sin(pi-theta) = sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2}+thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2} -thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi+theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi - theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2}-thetabigr)=sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2} + thetabigr)= -sintheta ; qquad 0 leq theta leq frac{pi}{2}$
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
$endgroup$
Because
$e^{itheta}=costheta + icdot sintheta$,
$e^{-itheta}=cos(-theta)+icdot sin(-theta)=costheta - i sin(theta)$.
$sin(2pi+t)=sin(t)$, same for cosine also.
$sinleft(frac{9pi}{4}right)=sin left(2pi + frac{pi}{4}right)$.
So that
begin{align*}
e^{-ifrac{9pi}{4}}
&=cosfrac{-9pi}{4}+i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} +i sinfrac{-9pi}{4}\
&=cosfrac{9pi}{4} -isinfrac{9pi}{4}\
&= cosleft(2pi + frac{pi}{4}right)-isinleft(2pi +frac{pi}{4}right)\
&=cosfrac{pi}{4}-isinfrac{pi}{4}\
&=e^{-ifrac{pi}{4}}
end{align*}
Also while doing such problems, these formulae may come handy:
$sin(pi+theta)= - sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sin(pi-theta) = sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2}+thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$sinbigl(frac{pi}{2} -thetabigr) = costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi+theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cos(pi - theta)=-costheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2}-thetabigr)=sintheta ; qquad 0 leq theta leq frac{pi}{2}$
$cosbigl(frac{pi}{2} + thetabigr)= -sintheta ; qquad 0 leq theta leq frac{pi}{2}$
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
edited Dec 11 '18 at 8:04
Rócherz
3,0013821
3,0013821
answered May 23 '11 at 14:10
user9413
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
|
show 4 more comments
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
4
4
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
So is $e^{-i5pi/4}$ also equal to $e^{-ipi/4}$, because $5 equiv 1 (text{mod} 4)$?
$endgroup$
– Jonas Meyer
May 23 '11 at 14:17
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.
$endgroup$
– user9413
May 23 '11 at 14:41
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.
$endgroup$
– Dooodle
May 23 '11 at 14:44
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5pi/4 = pi + pi(4)$. So $sin$ becomes negative and cos becomes positive.
$endgroup$
– user9413
May 23 '11 at 14:47
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
$begingroup$
@Doodle: $e^{-5pi/4}= cos(-5pi/4) +isin(-5pi/4) =cos(5pi/4)-isin(5pi/4) = cosfrac{pi}{4} +isinfrac{pi}{4}$
$endgroup$
– user9413
May 23 '11 at 14:49
|
show 4 more comments
$begingroup$
Note that $9pi/4 = (1+8)left(pi/4right) = pi/4 + 2pi$ and $e^{2pi i}=1$.
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Note that $9pi/4 = (1+8)left(pi/4right) = pi/4 + 2pi$ and $e^{2pi i}=1$.
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Note that $9pi/4 = (1+8)left(pi/4right) = pi/4 + 2pi$ and $e^{2pi i}=1$.
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
$endgroup$
Note that $9pi/4 = (1+8)left(pi/4right) = pi/4 + 2pi$ and $e^{2pi i}=1$.
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
edited May 23 '11 at 14:23
Eric Naslund
60.7k10140242
60.7k10140242
answered May 23 '11 at 14:14
lhflhf
167k11172403
answered May 23 '11 at 14:14
lhflhf
167k11172403
answered May 23 '11 at 14:14
lhflhf
167k11172403
167k11172403
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f40835%2fcomplex-exponential-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
$begingroup$
Poles? $ $ $ $ $ $
$endgroup$
– Did
May 23 '11 at 14:10