complex exponential equation $e^jzpi+j=0$
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
Which are the solutions of this equation?
$e^jzpi+j=0$
i was using the formula
$cos(...)+jsin(...)$
but it gives wrong results... I can't understand how to get it step by step
complex-analysis complex-numbers
edited Aug 28 at 13:21
amWhy
190k26221433
asked Aug 28 at 13:05
xmaionx
44
add a comment |Â
up vote
-2
down vote
favorite
Which are the solutions of this equation?
$e^jzpi+j=0$
i was using the formula
$cos(...)+jsin(...)$
but it gives wrong results... I can't understand how to get it step by step
complex-analysis complex-numbers
edited Aug 28 at 13:21
amWhy
190k26221433
asked Aug 28 at 13:05
xmaionx
44
2
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
2
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Which are the solutions of this equation?
$e^jzpi+j=0$
i was using the formula
$cos(...)+jsin(...)$
but it gives wrong results... I can't understand how to get it step by step
complex-analysis complex-numbers
edited Aug 28 at 13:21
amWhy
190k26221433
asked Aug 28 at 13:05
xmaionx
44
Which are the solutions of this equation?
$e^jzpi+j=0$
i was using the formula
$cos(...)+jsin(...)$
but it gives wrong results... I can't understand how to get it step by step
complex-analysis complex-numbers
edited Aug 28 at 13:21
amWhy
190k26221433
asked Aug 28 at 13:05
xmaionx
44
edited Aug 28 at 13:21
amWhy
190k26221433
edited Aug 28 at 13:21
amWhy
190k26221433
edited Aug 28 at 13:21
amWhy
190k26221433
190k26221433
asked Aug 28 at 13:05
xmaionx
44
asked Aug 28 at 13:05
xmaionx
44
asked Aug 28 at 13:05
xmaionx
44
44
2
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
2
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30
add a comment |Â
2
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
2
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30
2
2
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
2
2
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Recall that
$$-j=e^jleft(frac3pi2+2kpiright)$$
therefore
$$e^jzpi+j=0 iff e^jzpi=-j=e^jleft(frac3pi2+2kpiright)$$
that is
$$zpi=frac3pi2+2kpi implies z=frac32+2k quad kin mathbbZ$$
answered Aug 28 at 13:13
gimusi
70.9k73786
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
add a comment |Â
up vote
1
down vote
You have to solve $e^jzpi=-j$. Now $-j=e^-jpi/2$ and $e^a=e^b$ if and only if $a-b=2kpi j$ for an integer $k$. Can you take it from there?
answered Aug 28 at 13:13
Kusma
3,260218
add a comment |Â
up vote
1
down vote
You can use the fact that $e^a=e^b$ if and only if $a=b+2pi k i$ for integral $k$.
Put your equation in this form as follows:
$$e^izpi + i=0$$
$$e^izpi=-i$$
$$e^izpi=e^-pi i/2tagsince $-i$ can be written as $e^-pi i/2$$$
So this is true if and only if
$$izpi = -pi i/2 + 2pi ki = pi i(2k - tfrac12)$$
$$boxedz = 2k-tfrac12$$
for integral $k$. These are the values
$$leftldots, -frac92, -frac52,-frac12,frac32,frac72,frac112,ldots right$$
answered Aug 28 at 13:14
MPW
28.6k11853
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Recall that
$$-j=e^jleft(frac3pi2+2kpiright)$$
therefore
$$e^jzpi+j=0 iff e^jzpi=-j=e^jleft(frac3pi2+2kpiright)$$
that is
$$zpi=frac3pi2+2kpi implies z=frac32+2k quad kin mathbbZ$$
answered Aug 28 at 13:13
gimusi
70.9k73786
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
add a comment |Â
up vote
3
down vote
accepted
Recall that
$$-j=e^jleft(frac3pi2+2kpiright)$$
therefore
$$e^jzpi+j=0 iff e^jzpi=-j=e^jleft(frac3pi2+2kpiright)$$
that is
$$zpi=frac3pi2+2kpi implies z=frac32+2k quad kin mathbbZ$$
answered Aug 28 at 13:13
gimusi
70.9k73786
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Recall that
$$-j=e^jleft(frac3pi2+2kpiright)$$
therefore
$$e^jzpi+j=0 iff e^jzpi=-j=e^jleft(frac3pi2+2kpiright)$$
that is
$$zpi=frac3pi2+2kpi implies z=frac32+2k quad kin mathbbZ$$
answered Aug 28 at 13:13
gimusi
70.9k73786
Recall that
$$-j=e^jleft(frac3pi2+2kpiright)$$
therefore
$$e^jzpi+j=0 iff e^jzpi=-j=e^jleft(frac3pi2+2kpiright)$$
that is
$$zpi=frac3pi2+2kpi implies z=frac32+2k quad kin mathbbZ$$
answered Aug 28 at 13:13
gimusi
70.9k73786
edited Aug 28 at 13:20
answered Aug 28 at 13:13
gimusi
70.9k73786
answered Aug 28 at 13:13
gimusi
70.9k73786
answered Aug 28 at 13:13
gimusi
70.9k73786
70.9k73786
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
add a comment |Â
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
i get to that but this way don't you miss the -1/2 ? cause this is part from an integral and -1/2 it's a pole so this would change
– xmaionx
Aug 28 at 13:15
2
2
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
@xmaionx we dont loose solutions since for $k=-1$ we have $-frac12=frac32+2cdot (-1)$.
– gimusi
Aug 28 at 13:17
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
isn't k only positive numbers ?
– xmaionx
Aug 28 at 13:18
1
1
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
@xmaionx $k$ is an integer that is $kin mathbbZ=0,pm 1,pm 2, ldots$
– gimusi
Aug 28 at 13:19
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
Note that the numbers called $k$ in line two and line three are possibly different. (or the first and second lines are "for all $kinmathbbZ$" while the third is "there exists a $kinmathbbZ$"). It does not matter for the calculation here, though.
– Kusma
Aug 28 at 13:33
add a comment |Â
up vote
1
down vote
You have to solve $e^jzpi=-j$. Now $-j=e^-jpi/2$ and $e^a=e^b$ if and only if $a-b=2kpi j$ for an integer $k$. Can you take it from there?
answered Aug 28 at 13:13
Kusma
3,260218
add a comment |Â
up vote
1
down vote
You have to solve $e^jzpi=-j$. Now $-j=e^-jpi/2$ and $e^a=e^b$ if and only if $a-b=2kpi j$ for an integer $k$. Can you take it from there?
answered Aug 28 at 13:13
Kusma
3,260218
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have to solve $e^jzpi=-j$. Now $-j=e^-jpi/2$ and $e^a=e^b$ if and only if $a-b=2kpi j$ for an integer $k$. Can you take it from there?
answered Aug 28 at 13:13
Kusma
3,260218
You have to solve $e^jzpi=-j$. Now $-j=e^-jpi/2$ and $e^a=e^b$ if and only if $a-b=2kpi j$ for an integer $k$. Can you take it from there?
answered Aug 28 at 13:13
Kusma
3,260218
answered Aug 28 at 13:13
Kusma
3,260218
answered Aug 28 at 13:13
Kusma
3,260218
answered Aug 28 at 13:13
Kusma
3,260218
3,260218
add a comment |Â
add a comment |Â
up vote
1
down vote
You can use the fact that $e^a=e^b$ if and only if $a=b+2pi k i$ for integral $k$.
Put your equation in this form as follows:
$$e^izpi + i=0$$
$$e^izpi=-i$$
$$e^izpi=e^-pi i/2tagsince $-i$ can be written as $e^-pi i/2$$$
So this is true if and only if
$$izpi = -pi i/2 + 2pi ki = pi i(2k - tfrac12)$$
$$boxedz = 2k-tfrac12$$
for integral $k$. These are the values
$$leftldots, -frac92, -frac52,-frac12,frac32,frac72,frac112,ldots right$$
answered Aug 28 at 13:14
MPW
28.6k11853
add a comment |Â
up vote
1
down vote
You can use the fact that $e^a=e^b$ if and only if $a=b+2pi k i$ for integral $k$.
Put your equation in this form as follows:
$$e^izpi + i=0$$
$$e^izpi=-i$$
$$e^izpi=e^-pi i/2tagsince $-i$ can be written as $e^-pi i/2$$$
So this is true if and only if
$$izpi = -pi i/2 + 2pi ki = pi i(2k - tfrac12)$$
$$boxedz = 2k-tfrac12$$
for integral $k$. These are the values
$$leftldots, -frac92, -frac52,-frac12,frac32,frac72,frac112,ldots right$$
answered Aug 28 at 13:14
MPW
28.6k11853
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can use the fact that $e^a=e^b$ if and only if $a=b+2pi k i$ for integral $k$.
Put your equation in this form as follows:
$$e^izpi + i=0$$
$$e^izpi=-i$$
$$e^izpi=e^-pi i/2tagsince $-i$ can be written as $e^-pi i/2$$$
So this is true if and only if
$$izpi = -pi i/2 + 2pi ki = pi i(2k - tfrac12)$$
$$boxedz = 2k-tfrac12$$
for integral $k$. These are the values
$$leftldots, -frac92, -frac52,-frac12,frac32,frac72,frac112,ldots right$$
answered Aug 28 at 13:14
MPW
28.6k11853
You can use the fact that $e^a=e^b$ if and only if $a=b+2pi k i$ for integral $k$.
Put your equation in this form as follows:
$$e^izpi + i=0$$
$$e^izpi=-i$$
$$e^izpi=e^-pi i/2tagsince $-i$ can be written as $e^-pi i/2$$$
So this is true if and only if
$$izpi = -pi i/2 + 2pi ki = pi i(2k - tfrac12)$$
$$boxedz = 2k-tfrac12$$
for integral $k$. These are the values
$$leftldots, -frac92, -frac52,-frac12,frac32,frac72,frac112,ldots right$$
answered Aug 28 at 13:14
MPW
28.6k11853
answered Aug 28 at 13:14
MPW
28.6k11853
answered Aug 28 at 13:14
MPW
28.6k11853
answered Aug 28 at 13:14
MPW
28.6k11853
28.6k11853
add a comment |Â
add a comment |Â
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2
Can you edit your answer to show what you did with the formula to get the wrong results? That might help us get a better idea of where you went wrong.
– Logan Toll
Aug 28 at 13:12
i made similar steps to @gimuli but i get 3/2+2k , but this way i miss the -1/2 solution...
– xmaionx
Aug 28 at 13:17
@xmaionx Note that we have $-frac12+2k, frac32+2h, frac52+2m$ and so on are all equivalent solutions
– gimusi
Aug 28 at 13:18
2
@xmaionx show your work and effort in the OP in order to avoid downvotes
– gimusi
Aug 28 at 13:30