Solving complex number equations involving trig
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Let $w$ be a complex number. By solving the equation $frac{u^2 - 1}{u^2 + 1} = iw$ for a suitable complex number $u$, find an expression for $tan^{-1}(w)$.
I did the first part and got that $u^2 = frac{1 + iw}{1 - iw}$ but have no idea where to go from here. Should I be integrating or something?
complex-analysis complex-numbers
asked Nov 18 at 0:57
sam
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up vote
0
down vote
favorite
Let $w$ be a complex number. By solving the equation $frac{u^2 - 1}{u^2 + 1} = iw$ for a suitable complex number $u$, find an expression for $tan^{-1}(w)$.
I did the first part and got that $u^2 = frac{1 + iw}{1 - iw}$ but have no idea where to go from here. Should I be integrating or something?
complex-analysis complex-numbers
asked Nov 18 at 0:57
sam
498
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $w$ be a complex number. By solving the equation $frac{u^2 - 1}{u^2 + 1} = iw$ for a suitable complex number $u$, find an expression for $tan^{-1}(w)$.
I did the first part and got that $u^2 = frac{1 + iw}{1 - iw}$ but have no idea where to go from here. Should I be integrating or something?
complex-analysis complex-numbers
asked Nov 18 at 0:57
sam
498
Let $w$ be a complex number. By solving the equation $frac{u^2 - 1}{u^2 + 1} = iw$ for a suitable complex number $u$, find an expression for $tan^{-1}(w)$.
I did the first part and got that $u^2 = frac{1 + iw}{1 - iw}$ but have no idea where to go from here. Should I be integrating or something?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 18 at 0:57
sam
498
asked Nov 18 at 0:57
sam
498
edited Nov 18 at 1:09
asked Nov 18 at 0:57
sam
498
asked Nov 18 at 0:57
sam
498
asked Nov 18 at 0:57
sam
498
498
add a comment |
add a comment |
2 Answers
2
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3
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This is kind of a neat one.
$$ frac{u^2 - 1}{u^2 + 1} = iw $$
$$ frac{ frac{u - 1/u }{2i} }{ frac{u + 1/u }{2} }= w $$
The definition of sine and cosine can be used with the right choice of $u$.
$$ u = e^{itheta} $$
$$ frac{ frac{e^{itheta} - e^{-itheta} }{2i} }{ frac{e^{itheta} + e^{-itheta} }{2} } = w $$
$$ frac{ sin(theta)} {cos(theta)} = tan(theta)= w $$
Finally, the expression can be converted back into terms of $u$.
$$ tan^{-1}(w)= theta = frac{ln(u)}{i} = -ln(u)i $$
answered Nov 18 at 1:34
Cedron Dawg
80019
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Let $gamma=tan ^{-1} w$ then we have $$w=tan gamma =(1/i)(frac {e^{2i gamma }-1}{e^{2i gamma}+1}) $$
Thus with $u=e^{igamma} $we have $$ tan ^{-1} w=frac {ln u}{i}$$
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is kind of a neat one.
$$ frac{u^2 - 1}{u^2 + 1} = iw $$
$$ frac{ frac{u - 1/u }{2i} }{ frac{u + 1/u }{2} }= w $$
The definition of sine and cosine can be used with the right choice of $u$.
$$ u = e^{itheta} $$
$$ frac{ frac{e^{itheta} - e^{-itheta} }{2i} }{ frac{e^{itheta} + e^{-itheta} }{2} } = w $$
$$ frac{ sin(theta)} {cos(theta)} = tan(theta)= w $$
Finally, the expression can be converted back into terms of $u$.
$$ tan^{-1}(w)= theta = frac{ln(u)}{i} = -ln(u)i $$
answered Nov 18 at 1:34
Cedron Dawg
80019
add a comment |
up vote
3
down vote
accepted
This is kind of a neat one.
$$ frac{u^2 - 1}{u^2 + 1} = iw $$
$$ frac{ frac{u - 1/u }{2i} }{ frac{u + 1/u }{2} }= w $$
The definition of sine and cosine can be used with the right choice of $u$.
$$ u = e^{itheta} $$
$$ frac{ frac{e^{itheta} - e^{-itheta} }{2i} }{ frac{e^{itheta} + e^{-itheta} }{2} } = w $$
$$ frac{ sin(theta)} {cos(theta)} = tan(theta)= w $$
Finally, the expression can be converted back into terms of $u$.
$$ tan^{-1}(w)= theta = frac{ln(u)}{i} = -ln(u)i $$
answered Nov 18 at 1:34
Cedron Dawg
80019
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is kind of a neat one.
$$ frac{u^2 - 1}{u^2 + 1} = iw $$
$$ frac{ frac{u - 1/u }{2i} }{ frac{u + 1/u }{2} }= w $$
The definition of sine and cosine can be used with the right choice of $u$.
$$ u = e^{itheta} $$
$$ frac{ frac{e^{itheta} - e^{-itheta} }{2i} }{ frac{e^{itheta} + e^{-itheta} }{2} } = w $$
$$ frac{ sin(theta)} {cos(theta)} = tan(theta)= w $$
Finally, the expression can be converted back into terms of $u$.
$$ tan^{-1}(w)= theta = frac{ln(u)}{i} = -ln(u)i $$
answered Nov 18 at 1:34
Cedron Dawg
80019
This is kind of a neat one.
$$ frac{u^2 - 1}{u^2 + 1} = iw $$
$$ frac{ frac{u - 1/u }{2i} }{ frac{u + 1/u }{2} }= w $$
The definition of sine and cosine can be used with the right choice of $u$.
$$ u = e^{itheta} $$
$$ frac{ frac{e^{itheta} - e^{-itheta} }{2i} }{ frac{e^{itheta} + e^{-itheta} }{2} } = w $$
$$ frac{ sin(theta)} {cos(theta)} = tan(theta)= w $$
Finally, the expression can be converted back into terms of $u$.
$$ tan^{-1}(w)= theta = frac{ln(u)}{i} = -ln(u)i $$
answered Nov 18 at 1:34
Cedron Dawg
80019
answered Nov 18 at 1:34
Cedron Dawg
80019
answered Nov 18 at 1:34
Cedron Dawg
80019
answered Nov 18 at 1:34
Cedron Dawg
80019
80019
add a comment |
add a comment |
up vote
0
down vote
Let $gamma=tan ^{-1} w$ then we have $$w=tan gamma =(1/i)(frac {e^{2i gamma }-1}{e^{2i gamma}+1}) $$
Thus with $u=e^{igamma} $we have $$ tan ^{-1} w=frac {ln u}{i}$$
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
Let $gamma=tan ^{-1} w$ then we have $$w=tan gamma =(1/i)(frac {e^{2i gamma }-1}{e^{2i gamma}+1}) $$
Thus with $u=e^{igamma} $we have $$ tan ^{-1} w=frac {ln u}{i}$$
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $gamma=tan ^{-1} w$ then we have $$w=tan gamma =(1/i)(frac {e^{2i gamma }-1}{e^{2i gamma}+1}) $$
Thus with $u=e^{igamma} $we have $$ tan ^{-1} w=frac {ln u}{i}$$
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
Let $gamma=tan ^{-1} w$ then we have $$w=tan gamma =(1/i)(frac {e^{2i gamma }-1}{e^{2i gamma}+1}) $$
Thus with $u=e^{igamma} $we have $$ tan ^{-1} w=frac {ln u}{i}$$
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
edited Nov 18 at 1:38
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 1:32
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
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