Simplify the expression $ left(displaystyle frac{1+i z}{1-iz}right)^i, i^2=-1. $ [closed]
$begingroup$
Simplify the expression
$$
left(frac{1+i z}{1-iz}right)^i, i^2=-1, z in mathbb{R}.
$$
Really I have no any good idea to do it.
combinatorics complex-numbers
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
$endgroup$
closed as off-topic by jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn Dec 20 '18 at 1:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Simplify the expression
$$
left(frac{1+i z}{1-iz}right)^i, i^2=-1, z in mathbb{R}.
$$
Really I have no any good idea to do it.
combinatorics complex-numbers
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
$endgroup$
closed as off-topic by jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn Dec 20 '18 at 1:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
1
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41
add a comment |
$begingroup$
Simplify the expression
$$
left(frac{1+i z}{1-iz}right)^i, i^2=-1, z in mathbb{R}.
$$
Really I have no any good idea to do it.
combinatorics complex-numbers
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
$endgroup$
Simplify the expression
$$
left(frac{1+i z}{1-iz}right)^i, i^2=-1, z in mathbb{R}.
$$
Really I have no any good idea to do it.
combinatorics complex-numbers
combinatorics complex-numbers
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
asked Dec 19 '18 at 20:03
LeoxLeox
5,3431424
5,3431424
closed as off-topic by jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn Dec 20 '18 at 1:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn Dec 20 '18 at 1:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jameselmore, Davide Giraudo, Shailesh, Will Fisher, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
1
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41
add a comment |
1
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
1
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41
1
1
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
1
1
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Converting to polar form, we have
$$1 +iz = sqrt{1+z^2} e^{i(arctan z + 2npi)}, 1-iz = sqrt{1+z^2}e^{-i(arctan z + 2mpi)}$$
where $n, m$ can be any integers due to the periodicity of the complex exponential.
Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:
begin{align}
left(frac{1+iz}{1-iz}right)^i &= left(frac{e^{iarctan z}e^{2npi i}}{e^{-iarctan z}e^{2mpi i}}right)^i\
&=left(e^{2iarctan z}e^{2ipi k}right)^i\
&=e^{-2(arctan z +pi k)}
end{align}
where $k = n - m$ is also an integer.
answered Dec 19 '18 at 20:14
TomTom
2,791315
$endgroup$
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
add a comment |
$begingroup$
Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-
Rationalizing said expression
$=(frac{1-z^2+2iz}{1-z^2})^i=(1+ifrac{2z}{1-z^2})^i=r(e^{itheta})^i=re^{i^2theta}=re^{-theta}$
$$where, r=frac{1+z^2}{1-z^2} and tantheta=frac{2z}{1-z^2} $$
answered Dec 19 '18 at 20:31
MustangMustang
3367
$endgroup$
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Converting to polar form, we have
$$1 +iz = sqrt{1+z^2} e^{i(arctan z + 2npi)}, 1-iz = sqrt{1+z^2}e^{-i(arctan z + 2mpi)}$$
where $n, m$ can be any integers due to the periodicity of the complex exponential.
Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:
begin{align}
left(frac{1+iz}{1-iz}right)^i &= left(frac{e^{iarctan z}e^{2npi i}}{e^{-iarctan z}e^{2mpi i}}right)^i\
&=left(e^{2iarctan z}e^{2ipi k}right)^i\
&=e^{-2(arctan z +pi k)}
end{align}
where $k = n - m$ is also an integer.
answered Dec 19 '18 at 20:14
TomTom
2,791315
$endgroup$
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
add a comment |
$begingroup$
Converting to polar form, we have
$$1 +iz = sqrt{1+z^2} e^{i(arctan z + 2npi)}, 1-iz = sqrt{1+z^2}e^{-i(arctan z + 2mpi)}$$
where $n, m$ can be any integers due to the periodicity of the complex exponential.
Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:
begin{align}
left(frac{1+iz}{1-iz}right)^i &= left(frac{e^{iarctan z}e^{2npi i}}{e^{-iarctan z}e^{2mpi i}}right)^i\
&=left(e^{2iarctan z}e^{2ipi k}right)^i\
&=e^{-2(arctan z +pi k)}
end{align}
where $k = n - m$ is also an integer.
answered Dec 19 '18 at 20:14
TomTom
2,791315
$endgroup$
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
add a comment |
$begingroup$
Converting to polar form, we have
$$1 +iz = sqrt{1+z^2} e^{i(arctan z + 2npi)}, 1-iz = sqrt{1+z^2}e^{-i(arctan z + 2mpi)}$$
where $n, m$ can be any integers due to the periodicity of the complex exponential.
Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:
begin{align}
left(frac{1+iz}{1-iz}right)^i &= left(frac{e^{iarctan z}e^{2npi i}}{e^{-iarctan z}e^{2mpi i}}right)^i\
&=left(e^{2iarctan z}e^{2ipi k}right)^i\
&=e^{-2(arctan z +pi k)}
end{align}
where $k = n - m$ is also an integer.
answered Dec 19 '18 at 20:14
TomTom
2,791315
$endgroup$
Converting to polar form, we have
$$1 +iz = sqrt{1+z^2} e^{i(arctan z + 2npi)}, 1-iz = sqrt{1+z^2}e^{-i(arctan z + 2mpi)}$$
where $n, m$ can be any integers due to the periodicity of the complex exponential.
Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:
begin{align}
left(frac{1+iz}{1-iz}right)^i &= left(frac{e^{iarctan z}e^{2npi i}}{e^{-iarctan z}e^{2mpi i}}right)^i\
&=left(e^{2iarctan z}e^{2ipi k}right)^i\
&=e^{-2(arctan z +pi k)}
end{align}
where $k = n - m$ is also an integer.
answered Dec 19 '18 at 20:14
TomTom
2,791315
edited Dec 19 '18 at 21:37
answered Dec 19 '18 at 20:14
TomTom
2,791315
answered Dec 19 '18 at 20:14
TomTom
2,791315
answered Dec 19 '18 at 20:14
TomTom
2,791315
2,791315
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
add a comment |
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
It would be nice not to occult the other solutions when you deal with complex exponential. i.e. factors of $e^{2kpi}$.
$endgroup$
– zwim
Dec 19 '18 at 21:14
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
That's a good point. I've modified my answer accordingly. Thanks for pointing it out!
$endgroup$
– Tom
Dec 19 '18 at 21:33
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
$begingroup$
@Tom, Thank you!
$endgroup$
– Leox
Dec 19 '18 at 21:55
add a comment |
$begingroup$
Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-
Rationalizing said expression
$=(frac{1-z^2+2iz}{1-z^2})^i=(1+ifrac{2z}{1-z^2})^i=r(e^{itheta})^i=re^{i^2theta}=re^{-theta}$
$$where, r=frac{1+z^2}{1-z^2} and tantheta=frac{2z}{1-z^2} $$
answered Dec 19 '18 at 20:31
MustangMustang
3367
$endgroup$
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
add a comment |
$begingroup$
Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-
Rationalizing said expression
$=(frac{1-z^2+2iz}{1-z^2})^i=(1+ifrac{2z}{1-z^2})^i=r(e^{itheta})^i=re^{i^2theta}=re^{-theta}$
$$where, r=frac{1+z^2}{1-z^2} and tantheta=frac{2z}{1-z^2} $$
answered Dec 19 '18 at 20:31
MustangMustang
3367
$endgroup$
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
add a comment |
$begingroup$
Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-
Rationalizing said expression
$=(frac{1-z^2+2iz}{1-z^2})^i=(1+ifrac{2z}{1-z^2})^i=r(e^{itheta})^i=re^{i^2theta}=re^{-theta}$
$$where, r=frac{1+z^2}{1-z^2} and tantheta=frac{2z}{1-z^2} $$
answered Dec 19 '18 at 20:31
MustangMustang
3367
$endgroup$
Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-
Rationalizing said expression
$=(frac{1-z^2+2iz}{1-z^2})^i=(1+ifrac{2z}{1-z^2})^i=r(e^{itheta})^i=re^{i^2theta}=re^{-theta}$
$$where, r=frac{1+z^2}{1-z^2} and tantheta=frac{2z}{1-z^2} $$
answered Dec 19 '18 at 20:31
MustangMustang
3367
answered Dec 19 '18 at 20:31
MustangMustang
3367
answered Dec 19 '18 at 20:31
MustangMustang
3367
answered Dec 19 '18 at 20:31
MustangMustang
3367
3367
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
add a comment |
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
$begingroup$
$(e^{itheta})^i=e^{i^2theta}$ requires some justification
$endgroup$
– Mustang
Dec 19 '18 at 20:34
add a comment |
1
$begingroup$
Here's a hint: try converting the numerator and denominator to exponential form. i.e $e^{i.theta}$
$endgroup$
– Ananthakrishna
Dec 19 '18 at 20:06
1
$begingroup$
How is "simplified" defined for you? Involving $arctan (z)$ or $exp(z)$ does not appear to be simpler than just what we have.
$endgroup$
– Dietrich Burde
Dec 19 '18 at 20:24
$begingroup$
choosing $z$ as a variable name in $mathbb R$ where there are complex numbers everywhere is a poor choice (or a sadistic one...).
$endgroup$
– zwim
Dec 19 '18 at 21:12
$begingroup$
zwim I agree :)
$endgroup$
– Leox
Dec 19 '18 at 21:41