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.







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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
If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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














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.







share|cite|improve this question













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
If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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












up vote
1
down vote

favorite









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.







share|cite|improve this question













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.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 21:53
























asked Jul 20 at 21:44









TheQuantumMan

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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
If this question can be reworded to fit the rules in the help center, please edit the question.




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
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 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 $y=cosz=frace^iz+e^-iz2$ 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










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.






share|cite|improve this answer






























    1 Answer
    1






    active

    oldest

    votes








    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.






    share|cite|improve this answer



























      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer















        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.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 20 at 22:31


























        answered Jul 20 at 22:24









        B. Goddard

        16.4k21238




        16.4k21238












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