e , π , i , − 1 e, \pi, i, - 1 and 2018

Calculus Level 3

The absolute value, magnitude or module of the complex number ( 2018 i ) − 2018 i (2018i)^{-2018i} can be written as ∣ ( 2018 i ) − 2018 i ∣ = e A π |(2018i)^{-2018i}| = e^{A \pi} being A A an integer number.

  • Enter A A .

Assumption.- i = − 1 = e i π 2 i = \sqrt{- 1} = e^{i \frac{\pi}{2}} .


The answer is 1009.

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

∣ ( 2018 i ) − 2018 i ∣ = ∣ e − 2018 i ⋅ ln ⁡ ( 2018 i ) ∣ = ∣ e − 2018 i ⋅ ( ln ⁡ ∣ 2018 ∣ + i π 2 ) ∣ = e 1009 π \displaystyle \huge |(2018i)^{-2018i}| = |e^{-2018i \cdot \ln (2018i)}| = |e^{-2018i \cdot (\ln |2018| + i \frac{\pi}{2})}| = e^{1009 \pi}

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