anonymous
  • anonymous
Show that for any complex number Z Log(e^Z)=Z
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i meant Log(e^Z)=Z
anonymous
  • anonymous
yea, Log=In
anonymous
  • anonymous
That is true if we dealing with real numbers,but this is a complex Log/In where Z=x+iy

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More answers

perl
  • perl
log e^z = log e^(x+iy) = log e^x *e^(iy) = log e^x + log e^iy = x + iy= z
perl
  • perl
I think the logic is correct, not sure
experimentX
  • experimentX
log e^iy itself is a complex number with real part zero.
perl
  • perl
oh i got something weird ln ( e ^ ( 3 + 6i ) ) = 3 - .283185i
anonymous
  • anonymous
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)
perl
  • perl
i dont think this is true always
anonymous
  • anonymous
Here is the definition of LogZ=In[z]+iArg(z) where Arg -pie
perl
  • perl
so you are using the principal domain
anonymous
  • anonymous
yea
experimentX
  • experimentX
ln e^x e^yj = ln e^x + ln e^yj = x + ln (cos y + i sin y)
perl
  • perl
what is ln [z] , is that the real part of z ? [z]
perl
  • perl
or the modulus of z ?
anonymous
  • anonymous
Modulus of z
perl
  • perl
http://upload.wikimedia.org/wikipedia/en/math/7/5/b/75be122d4e8017a1259a34a3d509ba0d.png
anonymous
  • anonymous
experimentX, your method may work,,let me see
perl
  • perl
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
anonymous
  • anonymous
Thank you very much, i got it now.
experimentX
  • experimentX
http://upload.wikimedia.org/wikipedia/en/math/7/5/b/75be122d4e8017a1259a34a3d509ba0d.png
anonymous
  • anonymous
yea, that is a great clue.thanks
experimentX
  • experimentX
@answer you still there ... ??
anonymous
  • anonymous
I went for a break, i'm back now.
experimentX
  • experimentX
from this ln (cos y + i sin y) = i y I might have been wrong here !! still it gives ln e^Z = z ... sorry

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