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Find All Solutions E Z 1

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E^z = -1 -i (find all z)

  • Thread starter vilhelm
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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]

Answers and Replies

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.

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.

[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]

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.
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 ?

No, b = 5 π /4 leads to sin(b) = cos(b) = -1, so ea = 1
Careful; both sin and cos are positive in the first quadrant. Also, check the periodicity.
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 ?


The exponential has an imaginary period of 2pi. e^z has no real period, period.
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 ?
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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Find All Solutions E Z 1

Source: https://www.physicsforums.com/threads/e-z-1-i-find-all-z.542061/