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If z=(i)^{(i)^{(i)}} where i= {-1}, then |z| is equal to

Question

If $z = (i)^{(i)^{(i)}}$ where $i = - 1$ , then |z| is equal to

e^{- π /2}

e^{- π}

None of these

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Owen Discussed

z = (i)^{(i)^{(i)}}

where

i = - 1

, then |z| is equal to

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

l o g (z) = i^{i} l o g (i)

= e^{\frac{iπ}{2} . i} l o g (e^{\frac{iπ}{2}})

= e^{\frac{- π}{2}} i \frac{π}{2}

Now

e^{\frac{- π}{2}} (\frac{π}{2})

is constant.
Therefore

e^{\frac{- π}{2}} (\frac{π}{2}) = λ

l o g (z) = λ (i)

z = e^{λi}

\Rightarrow ∣ z ∣ = 1

Hence, option 'A' is correct.

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Question Text	If $z = (i)^{(i)^{(i)}}$ where $i = - 1$ , then \|z\| is equal to
Answer Type	Text solution:1
Upvotes	150