I need help proving $operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 '18 at 9:42
Bernard
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
add a comment |
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 '18 at 9:42
Bernard
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
add a comment |
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
edited Nov 29 '18 at 9:42
Bernard
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
This is an example from my textbook.
$$operatorname{Im}(ie^{-2t}(cos(2t)+isin(2t))=e^{-2t}cos(2t)$$
I don't understand why the imaginary part of this expression equals $e^{-2t}cos(2t)$
Can anyone clarify this?
complex-numbers
complex-numbers
edited Nov 29 '18 at 9:42
Bernard
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
edited Nov 29 '18 at 9:42
Bernard
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
edited Nov 29 '18 at 9:42
Bernard
118k639112
edited Nov 29 '18 at 9:42
Bernard
118k639112
edited Nov 29 '18 at 9:42
Bernard
118k639112
118k639112
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
asked Nov 29 '18 at 9:10
Boris GrunwaldBoris Grunwald
1487
1487
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
add a comment |
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
add a comment |
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
Recall that for a complex number $z=a+ib$, where $a,binmathbb{R}$, the number $a$ is said to be the real part and $b$ the imaginary part. In your case,
$$ ie^{-2t}(cos(2t)+isin(2t)) = ie^{-2t}cos(2t) + i^2e^{-2t}sin(2t) = underbrace{-e^{-2t}sin(2t)}_{text{real part}}+icdotunderbrace{e^{-2t}cos(2t)}_{text{imaginary part}}. $$
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
edited Nov 29 '18 at 9:14
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
answered Nov 29 '18 at 9:13
MisterRiemannMisterRiemann
5,8291624
5,8291624
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
add a comment |
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
But on the RHS the $i$ is gone. Why is that?
– Boris Grunwald
Nov 29 '18 at 9:14
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
@BorisGrunwald Recall that $i^2 = -1$.
– MisterRiemann
Nov 29 '18 at 9:15
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
Thanks, I get it now :)
– Boris Grunwald
Nov 29 '18 at 9:17
add a comment |
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