Question

Show that for any complex number Z

Log(e^Z)=Z

Answer 1

i meant Log(e^Z)=Z

Answer 2

yea, Log=In

Answer 3

That is true if we dealing with real numbers,but this is a complex Log/In where Z=x+iy

Answer 4

log e^z = log e^(x+iy) = log e^x *e^(iy) = log e^x + log e^iy = x + iy= z

Answer 5

I think the logic is correct, not sure

Answer 6

log e^iy itself is a complex number with real part zero.

Answer 7

oh i got something weird
ln ( e ^ ( 3 + 6i ) ) = 3 - .283185i

Answer 8

Let show you what i have got so far, Let z=x+iy-->Then Log(e^z)=Log(e^x+iy)-->Log(e^x e^iy)---> In[e^xe^iy]=+iArg(e^x e^iy)

Answer 9

i dont think this is true always

Answer 10

Here is the definition of LogZ=In[z]+iArg(z) where Arg -pie<theta<pie

Answer 11

so you are using the principal domain

Answer 12

yea

Answer 13

ln e^x e^yj = ln e^x + ln e^yj = x + ln (cos y + i sin y)

Answer 14

what is ln [z] , is that the real part of z ? [z]

Answer 15

or the modulus of z ?

Answer 16

Modulus of z

Answer 17

http://upload.wikimedia.org/wikipedia/en/math/7/5/b/75be122d4e8017a1259a34a3d509ba0d.png

Answer 18

experimentX, your method may work,,let me see

Answer 19

ln e^(x+iy) = ln [ e^x * e^yi)
now we know ln ( r e^(itheta) ) = ln r + i theta. so ...
ln [ e^x * e^yi) = ln (e^x) + i*y = x + iy = z

Answer 20

Thank you very much, i got it now.

Answer 21

http://upload.wikimedia.org/wikipedia/en/math/7/5/b/75be122d4e8017a1259a34a3d509ba0d.png

Answer 22

yea, that is a great clue.thanks

Answer 23

@answer you still there ... ??

Answer 24

I went for a break, i'm back now.

Answer 25

from this
ln (cos y + i sin y) = i y

I might have been wrong here !!

still it gives ln e^Z = z ... sorry