# Pre-interview_Test_Question_37

Whatâ€™s probably expected

Write i = e^{\frac \pi 2 i} then i^i = e^{\frac \pi 2 i^2} = e^{-\frac \pi 2}.

Discussion

This is an ill-defined expression as it stands, because we can write:

i = e^{\left(\frac \pi 2 + 2 \pi N\right)i}

for any integer N. Then using naive rules of exponentiation, we would apparently have:

i^i = e^{-\left(\frac \pi 2 + 2 \pi N\right)}

for any integer N. This shows that complex exponentiation is multivalued and we have to do some more work. We usually define a principal value for z^\alpha. We usually define z^\alpha = e^{\alpha \operatorname {Log} z} where \operatorname {Log} is a particular branch of the complex logarithm. (this too is multivalued, eg. note that for example 1 = e^0 = e^{2 \pi i}) Exponentiation of e to a complex power, of course, then being well-understood using Eulerâ€™s formula. Usually, we would define the principal logarithm by:

\operatorname {Log} (z) = \log |z| + i \arg(z)

where \arg is the argument of z with \arg(z) \in (-\pi, \pi] and \log is the usual real logarithm. You verify that e^{\operatorname {Log}(z)} = z. Using this logarithm and definition of z^\alpha, we indeed i^i = e^{i (\log 1 + i \arg i)} = e^{i \times \frac \pi 2 i} = e^{-\frac \pi 2} as we obtained from naive manipulations.

Sorry if this turned into a rant, itâ€™s a worthwhile discussion.

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This is great!!

Does this work:

e^(i\theta)= cos(\theta)+ isin(\theta)
Therefore: e^(i\pi/2)=i
i^i = (e^(i\pi/2))^i=e^(-\pi/2)

Or are some of the operations not allowed because of the complex numbers

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What you say is fine for this level hence â€śwhatâ€™s probably expectedâ€ť.

But the problem comes with the fact that you can also write e^{5 \pi i/2} = i â€śso thatâ€ť i^i = e^{-5 \pi/2}. Then we apparently have e^{-5 \pi/2} = e^{-\pi/2}.

You are right that this means that (z^a)^b = z^{ab} doesnâ€™t always apply for all complex numbers, depending how we actually define z^a with z and/or a (non-real) complex.

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