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

1. Oct 19, 2011

### 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.

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

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

2. Oct 19, 2011

### SammyS

Staff Emeritus
But, look just above this.

ea > 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.

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.

3. Oct 19, 2011

### vilhelm

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 ?

4. Oct 19, 2011

### SammyS

Staff Emeritus
No, b = 5π/4 leads to sin(b) = cos(b) = -1, so ea = 1

5. Oct 19, 2011

### Bacle2

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

6. Oct 19, 2011

### Bacle2

The exponential has an imaginary period of 2pi. e^z has no real period, period.

7. Oct 19, 2011

### vela

Staff Emeritus
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 $e^{z_n} = e^{z_0}e^{2\pi n i}=e^{z_0}$