Find all values and graph some of them in the complex plane: ln(-e^(-i))
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Recall the definition of the complex logarithm:
Ln(z) = ln|z| + i arg(z) + 2πik
...for some integer k. Fortunately, our z is already in polar form and we have:
|z| = -1
arg(z) = -1
Ln[-e^(-i)] = ln(-1) - i + 2πik
We can compute ln(-1) as follows:
-1 = 1e^[(i)(π)] <--- in polar form:
ln(-1) = ln{ e^[(i)(π)] }
ln(-1) = iπ
Ln[-e^(-i)] = iπ - i + 2πik
Ln[-e^(-i)] = (i)(π - 1 + 2πk)
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Now just sub in a few values for k.
k = 0 → (i)(π - 1)
k = 1 → (i)(3π - 1)
k = 2 → (i)(5π - 1)
Since these numbers are all purely imaginary and positive, they'll all be on the positive y axis. It's also evident that:
| (i)(π - 1 + 2πk) | = π - 1 + 2πk
So the only thing that's changing is the distance away from the origin, given by the expression above.
Done!
Ln(z) = ln|z| + i arg(z) + 2πik
...for some integer k. Fortunately, our z is already in polar form and we have:
|z| = -1
arg(z) = -1
Ln[-e^(-i)] = ln(-1) - i + 2πik
We can compute ln(-1) as follows:
-1 = 1e^[(i)(π)] <--- in polar form:
ln(-1) = ln{ e^[(i)(π)] }
ln(-1) = iπ
Ln[-e^(-i)] = iπ - i + 2πik
Ln[-e^(-i)] = (i)(π - 1 + 2πk)
********
Now just sub in a few values for k.
k = 0 → (i)(π - 1)
k = 1 → (i)(3π - 1)
k = 2 → (i)(5π - 1)
Since these numbers are all purely imaginary and positive, they'll all be on the positive y axis. It's also evident that:
| (i)(π - 1 + 2πk) | = π - 1 + 2πk
So the only thing that's changing is the distance away from the origin, given by the expression above.
Done!
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The polar form of a complex number is z = r e^(iθ). We have z = -e^(-i). Thus r = -1.
http://www.wolframalpha.com/in…
http://www.wolframalpha.com/in…
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