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Mathematics 13 Online
OpenStudy (anonymous):

What is the exact value of Ln I ?

OpenStudy (anonymous):

Is that supposed to be an "I" or a one?

OpenStudy (anonymous):

a I

OpenStudy (anonymous):

not a one :p

OpenStudy (anonymous):

If you know about the complex exponential it follows the following identity, because when dealing with complex and imaginary numbers the exponential and trig functions are closely tied to one another : \[e ^{i*x}=\cos(x)+i*\sin(x)\] where x is measured in radians... so: \[e ^{i*\frac{ \pi }{ 2 }}=\cos(\frac{ \pi }{ 2 }) + i*\sin(\frac{ \pi }{ 2 })=i\]So the natural logarithm of i must be: \[\ln(i) = i*\frac{ \pi }{ 2 } = \frac{ i*\pi }{ 2 }\]

OpenStudy (anonymous):

but the answers just say -1, 0, e, and 1

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