Applied And Computational Mathematics

Sunday, December 24, 2017

Why Does the natural logarithm of i equal to i pi /2



\( ln\left ( i \right ) = i\frac{\pi}{2} \)


To prove the upper identity, we aim to use the Euler's Formula for the complex numbers
 \(e^{i \theta} = cos(\theta) + isin(\theta)\). Since,  \( ln\left ( i \right ) = i\frac{\pi}{2} \) can be interpreted to \(e^{something} = i\) and this will prompt us to choose value of \(\theta\) that serve our need from Euler's Formula. So, by setting the value of \(\theta = \frac{\pi}{2}\), the right hand side of Euler's formula becomes:

            \(e^{i \frac{\pi}{2}} = cos(\frac{\pi}{2}) + isin(\frac{\pi}{2})\)

            \(\Rightarrow \) \(e^{i \frac{\pi}{2}} = 0 + i\left ( 1 \right )\)

Hence , \( ln\left ( i \right ) = i\frac{\pi}{2} \).







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