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Fast calc: i^i with Euler’s identity

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Introduction

I saw a long mathematical procedure to calculate i^i, but if we use Euler’s identity, it becomes really short and fast.

What would you learn?

  • Make use of Euler's identity to solve complex algebraic expressions

@Alfredo Maussa


Fast calc: i^i with Euler’s identity

17/04/2020

Recovered from Medium

I saw a long mathematical procedure to calculate i^i, but if we use Euler’s identity, it becomes really short and fast.

That is all, it is basically to clear -1, apply root square and power to i, which gives e^(-pi/2).