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Homework Help: E^z = -1 -i (find all z)

  1. Oct 19, 2011 #1
    The problem statement
    find all complex numbers z, that satisfies the equation
    [itex]e^{z}\; =\; -1\; -i[/itex]

    The attempt at a solution
    [itex]z=a+bi[/itex]

    [itex]e^{\left( a+bi \right)}\; =\; e^{a}\cdot e^{bi}\; =\; e^{a}\left( \cos b\; +\; i\; \sin b \right)[/itex]

    I seek
    [itex]\begin{cases}e^a cos(b)=-1 \\ e^a sin(b)=-1 \end{cases}[/itex]

    So, what b satisfies sin(b) = cos(b) ?
    In the unit circle, I see: [itex]b=\frac{\pi }{4}+\pi n[/itex] as that angle for b. b is done.

    [itex]e^{a}\cos \left( \frac{\pi }{4} \right)=-1[/itex]
    [itex]e^{a}=-\sqrt{2}[/itex]
    [itex]a=\ln \left( -\sqrt{2} \right)[/itex]

    [itex]\Rightarrow[/itex] [itex]z=\ln \left( -\sqrt{2} \right)+i\left( \frac{\pi }{4}+\pi n \right)[/itex]
     
  2. jcsd
  3. Oct 19, 2011 #2

    SammyS

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    But, look just above this.

    ea > 0, for all a. So if [itex]e^a \cos(b)=-1 \text{ and }e^a \sin(b)=-1\,,[/itex] then both sin(b) and cos(b) are negative, so the angle, b, must be in the third quadrant.

    Look at [itex]\displaystyle e^{a}\cos \left( \frac{5\pi }{4} \right)=-1\,.[/itex] Using the resulting value for a will fix your problem of having the logarithm of a negative number.
     
  4. Oct 19, 2011 #3
    And b=5π/4 leads to e^a=-1
    this means a=ln(√2)

    Thanks.

    But, what about finding ALL z? I struggle with this: e^z has 2πi, whilst the normal is 2π. Does that get compensated by the "i" in the exponential of e? So that 2π is the right answer: b=5π/4 + 2π·n ?
     
  5. Oct 19, 2011 #4

    SammyS

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    No, b = 5π/4 leads to sin(b) = cos(b) = -1, so ea = 1
     
  6. Oct 19, 2011 #5

    Bacle2

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    Careful; both sin and cos are positive in the first quadrant. Also, check the periodicity.
     
  7. Oct 19, 2011 #6

    Bacle2

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    The exponential has an imaginary period of 2pi. e^z has no real period, period.
     
  8. Oct 19, 2011 #7

    vela

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    Yes, that's right. Once you have one solution, namely [itex]z_0=\ln\sqrt{2}+i\frac{5\pi}{4}[/itex], the rest have the form [itex]z_n = z_0 +2\pi n i[/itex] because [itex]e^{z_n} = e^{z_0}e^{2\pi n i}=e^{z_0}[/itex]
     
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