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Find complex number z from equation

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find complex number z from equation
    Sinz+cosz=4


    2. Relevant equations



    3. The attempt at a solution
    I rewrite sinz+cosz=4 in form:
    [itex]\frac{(exp[i*z]-exp[-i*z])}{2i}[/itex]+[itex]\frac{(exp[i*z]+exp[-i*z])}{2}[/itex]=4
    then subsitute t=exp[i*z]

    So I gain t2(1+i)+8ti-1+i=0,I think i can rewrite it in form t2a+tb-c=0
    Next step could be to use Quadratic formula,but the roots are complicated, maybe I have mistaken ?





    so
     
    Last edited by a moderator: Dec 9, 2013
  2. jcsd
  3. Dec 9, 2013 #2

    haruspex

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    I get something a little different. Check your signs. If you still get the same answer, please post your working.

    Do you mean complex? Why would that be wrong? Post whatever you get.
     
  4. Dec 10, 2013 #3

    Yes,i got something different now.
    t2(1+i)-8ti+(1-i)=0

    so
    t1,2=-8i+(-)[itex]\sqrt{(8i)^2-4(1+i)(1-i)}[/itex]/(2(1+i))

    t1,2=-8i+(-)[itex]\sqrt{-64-8}[/itex]/(2(1+i))

    I ment complicated roots .
    I cat simplify them further.
     
  5. Dec 10, 2013 #4

    haruspex

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    You've changed two signs. If you'd only changed the first one we'd now agree. Please check again.
    Well, you can simplify -64-8, and get rid of the complex denominator. I see no difficulty in reducing it to the usual a+ib form. But actually you will want it in re form because you still have to find z.
     
  6. Dec 10, 2013 #5
    so finally i have t^2(1+i)-8ti+i-1=0

    Unfortunely i do not see how to reduce it. How do you recommend to get rid of complex denominator.
     
  7. Dec 10, 2013 #6

    haruspex

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    That's a standard procedure: multiply top and bottom by its conjugate.
     
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