Finding real and imaginary parts of $ -i ln(1pm sqrt{2})+{frac{pi}{2}}+2pi k$ [closed]
$begingroup$
I have found an expression for $arcsin(i)$ as a complex number $z$. It is
$$
z = -i ln(1pm sqrt{2})+{frac{pi}{2}}+2pi k
$$
How should I find the real and imaginary parts of $arcsin(i)$ from here?
complex-analysis
edited Dec 16 '18 at 12:54
Did
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
$endgroup$
closed as off-topic by Did, Eevee Trainer, Paul Frost, Lord_Farin, rtybase Dec 16 '18 at 22:58
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." – Eevee Trainer, Paul Frost, Lord_Farin, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have found an expression for $arcsin(i)$ as a complex number $z$. It is
$$
z = -i ln(1pm sqrt{2})+{frac{pi}{2}}+2pi k
$$
How should I find the real and imaginary parts of $arcsin(i)$ from here?
complex-analysis
edited Dec 16 '18 at 12:54
Did
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
$endgroup$
closed as off-topic by Did, Eevee Trainer, Paul Frost, Lord_Farin, rtybase Dec 16 '18 at 22:58
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." – Eevee Trainer, Paul Frost, Lord_Farin, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have found an expression for $arcsin(i)$ as a complex number $z$. It is
$$
z = -i ln(1pm sqrt{2})+{frac{pi}{2}}+2pi k
$$
How should I find the real and imaginary parts of $arcsin(i)$ from here?
complex-analysis
edited Dec 16 '18 at 12:54
Did
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
$endgroup$
I have found an expression for $arcsin(i)$ as a complex number $z$. It is
$$
z = -i ln(1pm sqrt{2})+{frac{pi}{2}}+2pi k
$$
How should I find the real and imaginary parts of $arcsin(i)$ from here?
complex-analysis
complex-analysis
edited Dec 16 '18 at 12:54
Did
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
edited Dec 16 '18 at 12:54
Did
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
edited Dec 16 '18 at 12:54
Did
247k23222458
edited Dec 16 '18 at 12:54
Did
247k23222458
edited Dec 16 '18 at 12:54
Did
247k23222458
247k23222458
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
asked Dec 16 '18 at 12:42
user3132457user3132457
1336
1336
closed as off-topic by Did, Eevee Trainer, Paul Frost, Lord_Farin, rtybase Dec 16 '18 at 22:58
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." – Eevee Trainer, Paul Frost, Lord_Farin, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Eevee Trainer, Paul Frost, Lord_Farin, rtybase Dec 16 '18 at 22:58
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." – Eevee Trainer, Paul Frost, Lord_Farin, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The real part is anything without a coefficient of $i$, and the imaginary part is anything with a coefficient of $i$. Note that for the imaginary part, you basically get rid of the $i$.
Some examples:
- Let $z = a+bi, a,b in mathbb{R}$. Then $Re(z) = a, Im(z) = b$.
- Let $z = 3 + sqrt{3} + isqrt2$. Then $Re(z) = 3 +sqrt{3}, Im(z) = sqrt2$.
- Let $z = e^{it}$, for $t in [0,2pi]$. Since $e^{it} = cos(t) + i sin(t)$, and for the $t$ given the $cos(t)$ and $sin(t)$ functions are real-valued, then $Re(z) = cos(t)$ and $Im(z) = sin(t)$.
Anecdotally, you might want to note that your $k$'s in your answer are integers. Just a small nitpick.
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The real part is anything without a coefficient of $i$, and the imaginary part is anything with a coefficient of $i$. Note that for the imaginary part, you basically get rid of the $i$.
Some examples:
- Let $z = a+bi, a,b in mathbb{R}$. Then $Re(z) = a, Im(z) = b$.
- Let $z = 3 + sqrt{3} + isqrt2$. Then $Re(z) = 3 +sqrt{3}, Im(z) = sqrt2$.
- Let $z = e^{it}$, for $t in [0,2pi]$. Since $e^{it} = cos(t) + i sin(t)$, and for the $t$ given the $cos(t)$ and $sin(t)$ functions are real-valued, then $Re(z) = cos(t)$ and $Im(z) = sin(t)$.
Anecdotally, you might want to note that your $k$'s in your answer are integers. Just a small nitpick.
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
add a comment |
$begingroup$
The real part is anything without a coefficient of $i$, and the imaginary part is anything with a coefficient of $i$. Note that for the imaginary part, you basically get rid of the $i$.
Some examples:
- Let $z = a+bi, a,b in mathbb{R}$. Then $Re(z) = a, Im(z) = b$.
- Let $z = 3 + sqrt{3} + isqrt2$. Then $Re(z) = 3 +sqrt{3}, Im(z) = sqrt2$.
- Let $z = e^{it}$, for $t in [0,2pi]$. Since $e^{it} = cos(t) + i sin(t)$, and for the $t$ given the $cos(t)$ and $sin(t)$ functions are real-valued, then $Re(z) = cos(t)$ and $Im(z) = sin(t)$.
Anecdotally, you might want to note that your $k$'s in your answer are integers. Just a small nitpick.
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
add a comment |
$begingroup$
The real part is anything without a coefficient of $i$, and the imaginary part is anything with a coefficient of $i$. Note that for the imaginary part, you basically get rid of the $i$.
Some examples:
- Let $z = a+bi, a,b in mathbb{R}$. Then $Re(z) = a, Im(z) = b$.
- Let $z = 3 + sqrt{3} + isqrt2$. Then $Re(z) = 3 +sqrt{3}, Im(z) = sqrt2$.
- Let $z = e^{it}$, for $t in [0,2pi]$. Since $e^{it} = cos(t) + i sin(t)$, and for the $t$ given the $cos(t)$ and $sin(t)$ functions are real-valued, then $Re(z) = cos(t)$ and $Im(z) = sin(t)$.
Anecdotally, you might want to note that your $k$'s in your answer are integers. Just a small nitpick.
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
$endgroup$
The real part is anything without a coefficient of $i$, and the imaginary part is anything with a coefficient of $i$. Note that for the imaginary part, you basically get rid of the $i$.
Some examples:
- Let $z = a+bi, a,b in mathbb{R}$. Then $Re(z) = a, Im(z) = b$.
- Let $z = 3 + sqrt{3} + isqrt2$. Then $Re(z) = 3 +sqrt{3}, Im(z) = sqrt2$.
- Let $z = e^{it}$, for $t in [0,2pi]$. Since $e^{it} = cos(t) + i sin(t)$, and for the $t$ given the $cos(t)$ and $sin(t)$ functions are real-valued, then $Re(z) = cos(t)$ and $Im(z) = sin(t)$.
Anecdotally, you might want to note that your $k$'s in your answer are integers. Just a small nitpick.
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
edited Dec 16 '18 at 13:16
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
answered Dec 16 '18 at 12:49
Eevee TrainerEevee Trainer
5,4791936
5,4791936
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
add a comment |
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
I know what are real and imaginary parts, but if I go with that knowledge, then I'll have $Re (z)={frac{pi}{2}}+2pi k, Im (z)=ln (1pm sqrt{2})$. But that's not correct.
$endgroup$
– user3132457
Dec 16 '18 at 14:32
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
The $ln()$ function would be negative. If that's still not correct, that suggests your calculations leading up to your calculation of $z$ are not correct.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 20:38
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
$begingroup$
That's correct actually. The answer in book is $Re (z)={frac{pi}{2}}+2pi k-iln ({sqrt{2}}+1), Im (z)=-{frac{pi}{2}}+2pi k-iln ({sqrt{2}}-1)$
$endgroup$
– user3132457
Dec 17 '18 at 17:53
add a comment |
