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Why does $e^{i\\pi}=-1$? - Mathematics Stack Exchange Euler's formula describes two equivalent ways to move in a circle Starting at the number 1 1, see multiplication as a transformation that changes the number 1 ⋅eiπ 1 ⋅ e i π Regular exponential growth continuously increases 1 1 by some rate; imaginary exponential growth continuously rotates a number in the complex plane Growing for π π units of time means going πradians π r a d i
Prove that $e^{i\\pi} = -1$ - Mathematics Stack Exchange When I first found out that eiπ = −1 e i π = 1, I was blown away Does anyone here know one of (many I'm sure) proofs of this phenomenal equation? I can perform all of the algebra to get the −1 1 But, where does this come from? What is the derivation?
Prove that $i^i$ is a real number - Mathematics Stack Exchange Here's a proof that I absolutely do not believe: take its complex conjugate, which is (i¯)i¯ = (1 i)−i =ii (i) i = (1 i) i = i i Since complex conjugation leaves it fixed, it’s real! EDIT: In answer to @Isaac’s comment, I think that to justify the formula above, you have to go through exactly the same arguments that most of the other answerers did For complex numbers u u and v v
How does $e^{\\pi i}$ equal $-1$ - Mathematics Stack Exchange @ColeJohnson, eiπ = −1 e i π = 1 is a consequence of how the objects e e, π π and raising a number to the power of a complex number are defined There is no deep truth behind the equation as some like to think You are confused because you have a feeling for what e e, π π and −1 1 are but you can't see the relation The mystery lies in the definition of i i and what it means to have
Conjugate of exponential imaginary number - Mathematics Stack Exchange You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
How to prove Eulers formula: $e^{it}=\\cos t +i\\sin t$? Actually, it is common to define eit e i t using your equation If something is to be proved we must start by asking what we know about the involved parameters, so how is your definition of eit e i t? Do you use a series or some other limit process?
Why $e^{i(π 3)} \\ne -1$? - Mathematics Stack Exchange I understand why eiπ = −1 e i π = 1 and as a result ei2π =(eiπ)2 = 1 e i 2 π = (e i π) 2 = 1 These results can be confirmed using Euler's formula But why does eiπ 3 ≠ −1 e i π 3 ≠ 1 as we can write it (eiπ)1 3 (e i π) 1 3 and as such (−1)1 3 (1) 1 3 the cubic root of −1 1 is −1 1 itself Yet using Euler's formula we get eiπ 3 = 12 + i 3√ 2 e i π 3 = 1 2
How can $i^i = e^{-\\pi 2}$ - Mathematics Stack Exchange I was asked a homework question: find ii i i The solution provided was as follows: Let A = ii A = i i log A = i log i log A = i log i Now, log i = logeiπ 2 = iπ2 log i = log e i π 2 = i π 2 So, log A = −π2 log A = π 2 Thus, ii =e−π 2 i i = e π 2 I understood how the result was obtained, but it is illogical I understand that multiplying by i i is equivalent to rotating the