What is the solution for e^z = -2?

[thomas49th]

Homework Statement

e^z = -2

Homework Equations

The Attempt at a Solution

Let z = a + ib
square it first

e^z = -2
e^2z = 4
2z = ln4
z = ln2

but I cannot go anywhere from there. Is what I've done mathematically sound?

Thanks
Thomas

 

[tiny-tim]

Hi Thomas!

(try using the X² icon just above the Reply box )

Why are you making it so complicated?

Hint: what's e^a? what's e^ib?

 

[thomas49th]

Hi Thomas!

(try using the X² icon just above the Reply box )

Why are you making it so complicated?

Hint: what's e^a? what's e^ib?

heheh

e^{a}e^{ib} = e^{a+ib}

Clever stuff

but while e^{a} is just e^{a}. e^{ib} = \cos b + i\sin b

So where next?

Thanks
Thomas

 

[tiny-tim]

hi thomas49th!

(just got up :zzz: …)

Next is e^acosb + ie^asinb = -2,

so b = … ?

 

[thomas49th]

Hi Tim,

I also, just got up :D

Even after 9 hours sleep I cannot see the easy step that must be around what you're hinting at.

I'm going to write this out

e to the a multiplied by cosine of b plus e to the a multiplied by sine of b i gives -2.

Ahh that didn't help. What the bloody hell is it! Good start to the day :(

 

[Office_Shredder]

Match up the real and imaginary parts

 

[tiny-tim]

Start with the imaginary part.

 

[thomas49th]

You mean compare coeffs?

sin(b) = 0
cos(b) = -2/(e^a)

but we don't know the quantity a

 

[Office_Shredder]

sin(b)=0

Therefore b=?

 

[thomas49th]

b = 0, \pm \pi or \pm 2 \pi :)

cos(0) = \frac{-2}{e^{a}}
e^{a}=-2
a=ln(-2)

Same problem as before:(

Unless

cos(\pi) = \frac{-2}{e^{2}}
-e^{a} = -2
a = \ln(2)

but that seems a bit arbitrary

Thanks
Thomas

 

[Theaumasch]

It is arbitrary; b can be pi+k*2*pi, where k is an integer.

 

[Hurkyl]

Is what I've done mathematically sound?


e^2z = 4
2z = ln4

Is "ln" the ordinary real logarithm? Then this step was not sound, because you forgot all of the other solutions. (much like when you take the square root of both sides, you have to consider two possibilities for sign)

2z = ln4
z = ln2

Is "ln" the multi-valued complex logarithm? The identity you tried to use doesn't apply here.


When I took complex analysis, they taught us what the real and imaginary parts of the complex logarithm were:

\log z = \ln |z| + i \arg(z)​

along with the fact that if e^w = z, then w is one of the infinitely many values of \log z.

Though, the above facts directly using the ideas mentioned in the thread, if you do things carefully so as not to introduce spurious solutions or eliminate good ones. It's probably a good idea to have the ability to do this.

 

[jackmell]

\log z = \ln |z| + i \arg(z)

\log(z)=\ln|z|+i(\theta+2n\pi),\quad n=0,\pm 1,\pm 2,\cdots

The imaginary part is like a spiral-staircase that never ends, no basement, no top floor.

Also, to the thread author:

e^z=-2

That's just:

z=\log(-2)

but look at log(-2) "multiwise" in terms of the infinitely-valued complex logarithm:

\log(-2)=\ln(2)+i(\pi+2n\pi),\quad n=0,\pm 1, \pm 2, \cdots