Show $overline{e^{itheta}} = e^{-itheta}$ using trig identities
$begingroup$
My book claims we can verify this using "standard trigonometric identities." The notation of the right-hand side is throwing me off. Does that mean the complex conjugate of $e^{itheta}$ equals its multiplicative inverse, i.e. $frac{1}{e^{itheta}}$?
trigonometry complex-numbers notation exponential-function
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
$endgroup$
add a comment |
$begingroup$
My book claims we can verify this using "standard trigonometric identities." The notation of the right-hand side is throwing me off. Does that mean the complex conjugate of $e^{itheta}$ equals its multiplicative inverse, i.e. $frac{1}{e^{itheta}}$?
trigonometry complex-numbers notation exponential-function
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
$endgroup$
add a comment |
$begingroup$
My book claims we can verify this using "standard trigonometric identities." The notation of the right-hand side is throwing me off. Does that mean the complex conjugate of $e^{itheta}$ equals its multiplicative inverse, i.e. $frac{1}{e^{itheta}}$?
trigonometry complex-numbers notation exponential-function
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
$endgroup$
My book claims we can verify this using "standard trigonometric identities." The notation of the right-hand side is throwing me off. Does that mean the complex conjugate of $e^{itheta}$ equals its multiplicative inverse, i.e. $frac{1}{e^{itheta}}$?
trigonometry complex-numbers notation exponential-function
trigonometry complex-numbers notation exponential-function
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
edited Dec 16 '18 at 21:01
Eric Wofsey
186k14215342
186k14215342
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
asked Dec 8 '18 at 17:25
hiroshinhiroshin
666
666
add a comment |
add a comment |
2 Answers
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$begingroup$
Yes. You have
$$
e^{it}=cos t+isin t.
$$
And
$$
e^{-it}=frac1{e^{it}}=frac1{cos t+isin t}=frac{cos t-isin t}{cos^2t+sin^2t}=cos t-isin t=overline{e^{it}}.
$$
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
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add a comment |
$begingroup$
Recall that
$$ e^{itheta} = costheta+isintheta, quad text{so} quad overline{e^{itheta}} = costheta-isintheta = cos(-theta)+isin(-theta) = e^{i(-theta)} = e^{-itheta}, $$
since the cosine is even while the sine is odd.
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
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2 Answers
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2 Answers
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$begingroup$
Yes. You have
$$
e^{it}=cos t+isin t.
$$
And
$$
e^{-it}=frac1{e^{it}}=frac1{cos t+isin t}=frac{cos t-isin t}{cos^2t+sin^2t}=cos t-isin t=overline{e^{it}}.
$$
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
Yes. You have
$$
e^{it}=cos t+isin t.
$$
And
$$
e^{-it}=frac1{e^{it}}=frac1{cos t+isin t}=frac{cos t-isin t}{cos^2t+sin^2t}=cos t-isin t=overline{e^{it}}.
$$
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
Yes. You have
$$
e^{it}=cos t+isin t.
$$
And
$$
e^{-it}=frac1{e^{it}}=frac1{cos t+isin t}=frac{cos t-isin t}{cos^2t+sin^2t}=cos t-isin t=overline{e^{it}}.
$$
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
Yes. You have
$$
e^{it}=cos t+isin t.
$$
And
$$
e^{-it}=frac1{e^{it}}=frac1{cos t+isin t}=frac{cos t-isin t}{cos^2t+sin^2t}=cos t-isin t=overline{e^{it}}.
$$
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 8 '18 at 17:27
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
$begingroup$
Recall that
$$ e^{itheta} = costheta+isintheta, quad text{so} quad overline{e^{itheta}} = costheta-isintheta = cos(-theta)+isin(-theta) = e^{i(-theta)} = e^{-itheta}, $$
since the cosine is even while the sine is odd.
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
$endgroup$
add a comment |
$begingroup$
Recall that
$$ e^{itheta} = costheta+isintheta, quad text{so} quad overline{e^{itheta}} = costheta-isintheta = cos(-theta)+isin(-theta) = e^{i(-theta)} = e^{-itheta}, $$
since the cosine is even while the sine is odd.
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
$endgroup$
add a comment |
$begingroup$
Recall that
$$ e^{itheta} = costheta+isintheta, quad text{so} quad overline{e^{itheta}} = costheta-isintheta = cos(-theta)+isin(-theta) = e^{i(-theta)} = e^{-itheta}, $$
since the cosine is even while the sine is odd.
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
$endgroup$
Recall that
$$ e^{itheta} = costheta+isintheta, quad text{so} quad overline{e^{itheta}} = costheta-isintheta = cos(-theta)+isin(-theta) = e^{i(-theta)} = e^{-itheta}, $$
since the cosine is even while the sine is odd.
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
answered Dec 8 '18 at 17:28
MisterRiemannMisterRiemann
5,8021625
5,8021625
add a comment |
add a comment |
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