Identity regarding integral $frac{e^{iz}}{z}$
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Both R and r are real numbers. The path 1 is from -R to -r, the path 2 is from r to R. Why is this an identity when integral 1 is taken on path nr 1 and integral 2 on path nr 2 and integral 3 on path nr 2
$$intfrac{e^{iz}}{z}+ intfrac{e^{iz}}{z}=2iintfrac{sinz}{z}$$
complex-analysis complex-integration
asked Nov 22 at 16:24
ryszard eggink
1209
add a comment |
up vote
-1
down vote
favorite
Both R and r are real numbers. The path 1 is from -R to -r, the path 2 is from r to R. Why is this an identity when integral 1 is taken on path nr 1 and integral 2 on path nr 2 and integral 3 on path nr 2
$$intfrac{e^{iz}}{z}+ intfrac{e^{iz}}{z}=2iintfrac{sinz}{z}$$
complex-analysis complex-integration
asked Nov 22 at 16:24
ryszard eggink
1209
The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
1
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Both R and r are real numbers. The path 1 is from -R to -r, the path 2 is from r to R. Why is this an identity when integral 1 is taken on path nr 1 and integral 2 on path nr 2 and integral 3 on path nr 2
$$intfrac{e^{iz}}{z}+ intfrac{e^{iz}}{z}=2iintfrac{sinz}{z}$$
complex-analysis complex-integration
asked Nov 22 at 16:24
ryszard eggink
1209
Both R and r are real numbers. The path 1 is from -R to -r, the path 2 is from r to R. Why is this an identity when integral 1 is taken on path nr 1 and integral 2 on path nr 2 and integral 3 on path nr 2
$$intfrac{e^{iz}}{z}+ intfrac{e^{iz}}{z}=2iintfrac{sinz}{z}$$
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 22 at 16:24
ryszard eggink
1209
asked Nov 22 at 16:24
ryszard eggink
1209
edited Nov 22 at 16:29
asked Nov 22 at 16:24
ryszard eggink
1209
asked Nov 22 at 16:24
ryszard eggink
1209
asked Nov 22 at 16:24
ryszard eggink
1209
1209
The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
1
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35
add a comment |
The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
1
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35
The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
1
1
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35
add a comment |
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The identity is wrong. You may want to look up the Euler's formula for complex numbers.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:27
Sorry i was posting in on the phone , i missed the minus sign, i know the euler formula cosz+isinz , but how the cosine is reduced?
– ryszard eggink
Nov 22 at 16:29
1
I'm also commenting on phone. That doesn't stop you from adding dollar sign. Consider the formula for exp(-iz) and eliminate cos(z) from these two formulae. Make sin(z) the subject of the formula to conclude.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:31
why the harsh dislike mate?
– Aayush Paurana
Nov 22 at 16:32
@AayushPaurana The layout of this question is unclear. That's one of the reasons present in the tooltip text of the downvote button.
– GNUSupporter 8964民主女神 地下教會
Nov 22 at 16:35