Question

Find all complex numbers z∈C with |z| < 2π such that -e^z =e^-z.
Plot these points on the complex plane.

Answer 1

so far I got that |z| < 2π is a circle (x^2 + y^2 < 4π^2) but Im not sure what to do about the -e^z = e^-z bit...

Answer 2

@rational heya reckon you could help out?

Answer 3

\(\dfrac{e^z + e^{-z}}{2} = \cosh (z)\)

Answer 4

OHH yeahhh I forgot about that haha

Answer 5

That may bot be useful hmm

Answer 6

yea Im so lost

Answer 7

Lets try factoring maybe..

\( -e^{z} =e^{-z} \)

\( e^{z} +e^{-z}=0 \)

\( e^{-z}(e^{2z} +1)=0 \)

Answer 8

\(e^{-z}\) is never 0
so do we end up solving \(e^{2z}+1=0\) ?

Answer 9

hmm how do we do that though?

Answer 10

let \(e^z = t\)

Answer 11

ohh righttt

Answer 12

and then t = +-i?

Answer 13

Yep!

Answer 14

ohh so e^z = +-i

Answer 15

Yes, look at the argand plane

Answer 16

so z = +- π/2?

Answer 17

Nope, lets look at the argand plane

Answer 18

okay

Answer 19

is that the |z| < 2π part?

Answer 20

e^z = i

e^(x+iy) = i

e^x[cos(y) + isin(y)] = i

compare real parts both sides :
e^x cosy = 0

this gives x = 0, y = +- pi/2

therefore z = x + iy = 0+- ipi/2

Answer 21

ohh okay that makes sense

Answer 22

Notice, two solutions are +- ipi/2

not +- pi/2

Answer 23

yup

Answer 24

you should get two more solutions by solving

e^z = -i

Answer 25

would the two solutions just come out as +- πi/2 again?

Answer 26

http://www.wolframalpha.com/input/?i=solve+e%5E(2z)+%2B+1%3D0,+%7Cz%7C+%3C+2pi

Answer 27

but how do you get +-3πi/2 ?

Answer 28

Oh we're done actually, -pi/2 is same as 3pi/2

Answer 29

ohhh I get it, you just add 2π because it still lies in the |z| < 2

Answer 30

yeah lol thanks so much!! :D

Answer 31

the four solutions are then +-pi/2 and +-3pi/2

Answer 32

sorry I meant |z| < 2π *

Answer 33

yup thank you!!!

Answer 34

Np :)