Proof regarding Stein's and Sharachi's proof of the unit circle mapping onto the upper half plane - complex...
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I am currently working with Stein's and Shakarchi's Complex Analysis and trying to comprehend the proofs of certain theorems. But unfortunately i can't get behind the equation used in the proof of Theorem 1.2:
I understand why we are computing $Im(G(w))$ but i simply do not understand why it holds that $$Im(G(w)) = Re(G(u+iv))$$ even though i fully understand using $w = u+iv$ in order to compute both real part and imaginary part.
But the author simply computes the real part and claims that since $Re(G(u+iv)) > 0$ it holds that $Im(G(w)) > 0$ ?
I tried to solve it myself but i always end up having
$$Re(G(u+iv)) = frac{1-u^2-v^2}{(1-u)^2+v^2}$$ and $$Im(G(u+iv)) = frac{2v}{(1-u)^2+v^2}$$
and in order to proof we are mapping onto $mathbb{H}$ i would purely pay attention to the imaginary part, however, the author obviously uses the real part for his proof. What am i missing?
Thank you very much for any hint!
complex-analysis complex-geometry
asked Jan 12 at 23:37
ZestZest
287213
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add a comment |
$begingroup$
I am currently working with Stein's and Shakarchi's Complex Analysis and trying to comprehend the proofs of certain theorems. But unfortunately i can't get behind the equation used in the proof of Theorem 1.2:
I understand why we are computing $Im(G(w))$ but i simply do not understand why it holds that $$Im(G(w)) = Re(G(u+iv))$$ even though i fully understand using $w = u+iv$ in order to compute both real part and imaginary part.
But the author simply computes the real part and claims that since $Re(G(u+iv)) > 0$ it holds that $Im(G(w)) > 0$ ?
I tried to solve it myself but i always end up having
$$Re(G(u+iv)) = frac{1-u^2-v^2}{(1-u)^2+v^2}$$ and $$Im(G(u+iv)) = frac{2v}{(1-u)^2+v^2}$$
and in order to proof we are mapping onto $mathbb{H}$ i would purely pay attention to the imaginary part, however, the author obviously uses the real part for his proof. What am i missing?
Thank you very much for any hint!
complex-analysis complex-geometry
asked Jan 12 at 23:37
ZestZest
287213
$endgroup$
add a comment |
$begingroup$
I am currently working with Stein's and Shakarchi's Complex Analysis and trying to comprehend the proofs of certain theorems. But unfortunately i can't get behind the equation used in the proof of Theorem 1.2:
I understand why we are computing $Im(G(w))$ but i simply do not understand why it holds that $$Im(G(w)) = Re(G(u+iv))$$ even though i fully understand using $w = u+iv$ in order to compute both real part and imaginary part.
But the author simply computes the real part and claims that since $Re(G(u+iv)) > 0$ it holds that $Im(G(w)) > 0$ ?
I tried to solve it myself but i always end up having
$$Re(G(u+iv)) = frac{1-u^2-v^2}{(1-u)^2+v^2}$$ and $$Im(G(u+iv)) = frac{2v}{(1-u)^2+v^2}$$
and in order to proof we are mapping onto $mathbb{H}$ i would purely pay attention to the imaginary part, however, the author obviously uses the real part for his proof. What am i missing?
Thank you very much for any hint!
complex-analysis complex-geometry
asked Jan 12 at 23:37
ZestZest
287213
$endgroup$
I am currently working with Stein's and Shakarchi's Complex Analysis and trying to comprehend the proofs of certain theorems. But unfortunately i can't get behind the equation used in the proof of Theorem 1.2:
I understand why we are computing $Im(G(w))$ but i simply do not understand why it holds that $$Im(G(w)) = Re(G(u+iv))$$ even though i fully understand using $w = u+iv$ in order to compute both real part and imaginary part.
But the author simply computes the real part and claims that since $Re(G(u+iv)) > 0$ it holds that $Im(G(w)) > 0$ ?
I tried to solve it myself but i always end up having
$$Re(G(u+iv)) = frac{1-u^2-v^2}{(1-u)^2+v^2}$$ and $$Im(G(u+iv)) = frac{2v}{(1-u)^2+v^2}$$
and in order to proof we are mapping onto $mathbb{H}$ i would purely pay attention to the imaginary part, however, the author obviously uses the real part for his proof. What am i missing?
Thank you very much for any hint!
complex-analysis complex-geometry
complex-analysis complex-geometry
asked Jan 12 at 23:37
ZestZest
287213
asked Jan 12 at 23:37
ZestZest
287213
asked Jan 12 at 23:37
ZestZest
287213
asked Jan 12 at 23:37
ZestZest
287213
asked Jan 12 at 23:37
ZestZest
287213
287213
add a comment |
add a comment |
1 Answer
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He is just using the fact that $Im (i(a+ib))=Re (a+ib)$ for any $a,b in mathbb R$. Take $a+ib =frac {1-w} {1+w}$.
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
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Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
He is just using the fact that $Im (i(a+ib))=Re (a+ib)$ for any $a,b in mathbb R$. Take $a+ib =frac {1-w} {1+w}$.
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
add a comment |
$begingroup$
He is just using the fact that $Im (i(a+ib))=Re (a+ib)$ for any $a,b in mathbb R$. Take $a+ib =frac {1-w} {1+w}$.
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
add a comment |
$begingroup$
He is just using the fact that $Im (i(a+ib))=Re (a+ib)$ for any $a,b in mathbb R$. Take $a+ib =frac {1-w} {1+w}$.
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
He is just using the fact that $Im (i(a+ib))=Re (a+ib)$ for any $a,b in mathbb R$. Take $a+ib =frac {1-w} {1+w}$.
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 12 at 23:40
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
add a comment |
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
$begingroup$
Hello Kavi Rama Murthy, that's great. I didn't know that. Thank you very much for your help. I might return with a question tomorrow maybe, but for now i think i understood. thanks again!
$endgroup$
– Zest
Jan 12 at 23:52
add a comment |
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