Find expression for $arccos(z)$ with complex logarithm [closed]
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I want to find an expression for $arccos(z)$ with complex logarithm.
I know that it can be expressed as the exponential whose exponent is proportional to a complex logarithm, but I do not know how to do it.
complex-analysis complex-numbers logarithms
closed as off-topic by amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel Jul 22 at 13:33
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel
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I want to find an expression for $arccos(z)$ with complex logarithm.
I know that it can be expressed as the exponential whose exponent is proportional to a complex logarithm, but I do not know how to do it.
complex-analysis complex-numbers logarithms
closed as off-topic by amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel Jul 22 at 13:33
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel
Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13
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up vote
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up vote
1
down vote
favorite
I want to find an expression for $arccos(z)$ with complex logarithm.
I know that it can be expressed as the exponential whose exponent is proportional to a complex logarithm, but I do not know how to do it.
complex-analysis complex-numbers logarithms
I want to find an expression for $arccos(z)$ with complex logarithm.
I know that it can be expressed as the exponential whose exponent is proportional to a complex logarithm, but I do not know how to do it.
complex-analysis complex-numbers logarithms
edited Jul 20 at 21:53
asked Jul 20 at 21:44
TheQuantumMan
5831422
5831422
closed as off-topic by amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel Jul 22 at 13:33
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel
closed as off-topic by amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel Jul 22 at 13:33
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, Taroccoesbrocco, José Carlos Santos, Parcly Taxel
Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13
add a comment |Â
Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13
Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13
add a comment |Â
1 Answer
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If $w=arccos z,$ then $z=cos w = frace^iw+e^-iw2.$ We solve this equation for $w$:
$$2z = e^iw+e^-iw$$
$$2ze^iw = e^2iw+1$$
$$(e^iw)^2 -2ze^iw +1 = 0$$
$$e^iw = frac2zpm sqrt(2z)^2-42$$
$$e^iw = zpm sqrtz^2-1$$
$$iw = ln(zpm sqrtz^2-1)$$
$$w = fracln(zpm sqrtz^2-1)i.$$
Then think about the $pm$ sign.
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $w=arccos z,$ then $z=cos w = frace^iw+e^-iw2.$ We solve this equation for $w$:
$$2z = e^iw+e^-iw$$
$$2ze^iw = e^2iw+1$$
$$(e^iw)^2 -2ze^iw +1 = 0$$
$$e^iw = frac2zpm sqrt(2z)^2-42$$
$$e^iw = zpm sqrtz^2-1$$
$$iw = ln(zpm sqrtz^2-1)$$
$$w = fracln(zpm sqrtz^2-1)i.$$
Then think about the $pm$ sign.
add a comment |Â
up vote
1
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If $w=arccos z,$ then $z=cos w = frace^iw+e^-iw2.$ We solve this equation for $w$:
$$2z = e^iw+e^-iw$$
$$2ze^iw = e^2iw+1$$
$$(e^iw)^2 -2ze^iw +1 = 0$$
$$e^iw = frac2zpm sqrt(2z)^2-42$$
$$e^iw = zpm sqrtz^2-1$$
$$iw = ln(zpm sqrtz^2-1)$$
$$w = fracln(zpm sqrtz^2-1)i.$$
Then think about the $pm$ sign.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $w=arccos z,$ then $z=cos w = frace^iw+e^-iw2.$ We solve this equation for $w$:
$$2z = e^iw+e^-iw$$
$$2ze^iw = e^2iw+1$$
$$(e^iw)^2 -2ze^iw +1 = 0$$
$$e^iw = frac2zpm sqrt(2z)^2-42$$
$$e^iw = zpm sqrtz^2-1$$
$$iw = ln(zpm sqrtz^2-1)$$
$$w = fracln(zpm sqrtz^2-1)i.$$
Then think about the $pm$ sign.
If $w=arccos z,$ then $z=cos w = frace^iw+e^-iw2.$ We solve this equation for $w$:
$$2z = e^iw+e^-iw$$
$$2ze^iw = e^2iw+1$$
$$(e^iw)^2 -2ze^iw +1 = 0$$
$$e^iw = frac2zpm sqrt(2z)^2-42$$
$$e^iw = zpm sqrtz^2-1$$
$$iw = ln(zpm sqrtz^2-1)$$
$$w = fracln(zpm sqrtz^2-1)i.$$
Then think about the $pm$ sign.
edited Jul 20 at 22:31
answered Jul 20 at 22:24


B. Goddard
16.4k21238
16.4k21238
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Write $w=arccos z$, so $z=cos w$ and compute the real and imaginary parts.
– egreg
Jul 20 at 21:55
Write $y=cosz=frace^iz+e^-iz2$ and solve for $z$ (hint: its related to $arccosh$)
– aidangallagher4
Jul 20 at 22:13