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Homework Help: Complex Numbers and Euler's Identity

  1. May 14, 2017 #1
    1. The problem statement, all variables and given/known data
    exp(z)=-4+3i, find z in x+iy form

    2. Relevant equations
    See attached image.

    3. The attempt at a solution
    See attached image. exp(z)=exp(x+iy)=exp(x)*exp(iy)=exp(x)*[cos(y)+isin(y)] ... y=inv(tan(-3/4)=-.6432 .... mag(-4+3i)=5, x= ln (5)..exp(ln(5))=5 .... 5*[cos(-.6432)+isin(-6.432)] = -4+3i

    z=r*exp(i*theta)..z=5exp(-.6432) Did I do this right? File_000 (2).jpeg
  2. jcsd
  3. May 14, 2017 #2


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    2017 Award

    Staff: Mentor

    Check the value of the inverse tangent function you used. In particular, check the signs of the second last equation where you didn't see the error made earlier.

    The last line is wrong. You found z already, what do you do in the last line?
  4. May 14, 2017 #3
    you should take the lin of both sides its much simpler
    and use the the law for a lin of acomplex numper
  5. May 14, 2017 #4


    Staff: Mentor

    The natural log is usually abbreviated as "ln" not "lin".
  6. May 14, 2017 #5
    Theta should be the inv(tan(-3/4))= -.6435+pi =2.498 which fixes the sign error when I plug that in.

    The thing I am confused on is, is exp(ln(5))* [ cos(2.498)+isin(2.498) ] is that Z ? because that expression is equal to -4 +3i, but exp(-4+3i) ISNT equal to -4+3i
  7. May 14, 2017 #6
    I agree but the professor wants us to solve it using Euler's identities
  8. May 14, 2017 #7
    i know mark44 thanks . i saw some of my professors write it as lin.
  9. May 14, 2017 #8


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    Staff Emeritus
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    Hint: You found ##e^{\ln 5}e^{2.498i}=-4+3i = e^{z}##. Rewrite the left side in the form ##e^{x+iy}##.
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