How Do You Solve e^z = -1 - i for All Complex Numbers z?

[vilhelm]

The problem statement
find all complex numbers z, that satisfies the equation
e^{z}\; =\; -1\; -i

The attempt at a solution
z=a+bi

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

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

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

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

\Rightarrow z=\ln \left( -\sqrt{2} \right)+i\left( \frac{\pi }{4}+\pi n \right)

 

[SammyS]

The problem statement
find all complex numbers z, that satisfies the equation
e^{z}\; =\; -1\; -i

The attempt at a solution
z=a+bi

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

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

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

But, look just above this.

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

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

\Rightarrow z=\ln \left( -\sqrt{2} \right)+i\left( \frac{\pi }{4}+\pi n \right)

Look at \displaystyle e^{a}\cos \left( \frac{5\pi }{4} \right)=-1\,. Using the resulting value for a will fix your problem of having the logarithm of a negative number.

 

[vilhelm]

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 ?

 

[SammyS]

No, b = 5π/4 leads to sin(b) = cos(b) = -1, so e^a = 1

 

[Bacle2]

Careful; both sin and cos are positive in the first quadrant. Also, check the periodicity.

 

[Bacle2]

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.

 

[vela]

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 z_0=\ln\sqrt{2}+i\frac{5\pi}{4}, the rest have the form z_n = z_0 +2\pi n i because e^{z_n} = e^{z_0}e^{2\pi n i}=e^{z_0}