Can you spot that mistake?

Algebra Level 2

Dexter, a math lover, claims that he has proven e 2 i = 1 e^{2i}=1 , but applying the Euler's formula e i x = cos x + i sin x e^{ix}=\cos x + i \sin x , e 2 i e^{2i} is actually a complex number . In which of these steps does he first make a mistake?

Step 1 : As we know, e i π = 1 e^{i\pi}=-1 .
Step 2 : Squaring both sides, we get ( e i π ) 2 = 1 \big(e^{i\pi}\big)^2=1 .
Step 3 : Exchanging the exponents, ( e 2 i ) π = 1 \big(e^{2i}\big)^\pi=1 .
Step 4 : Multiplying 1 π \frac{1}{\pi} at the exponents of both sides, we claim that e 2 i = 1 e^{2i}=1 .

Step 1 Step 2 Step 3 Step 4 The proof is absolutely correct

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1 solution

Lee Dongheng
Jun 27, 2017

For equation x π = 1 x^{\pi}=1 where x C x\in\mathbb{C} , there are actually 3 roots which are x = 1 x=1 , x = e 2 i x=e^{2i} and x = e 2 i x=e^{-2i} . We cannot just conclude that e 2 i = 1 ! ! e^{2i}=1!!

Sir, how to find the roots of x π = 1 x^\pi=1 ?

Dexter Woo Teng Koon - 3 years, 11 months ago

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