e^(iπ ) = −1 squaring both sides e^(2πi) = 1 = e^0 comparing powers 2πi = 0 π = 0 or i = 0 ???


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Question Number 33473 by 33 last updated on 17/Apr/18

$e^{i π} = - 1$ $s q u a r i n g b o t h s i d e s$ $e^{2 π i} = 1 = e^{0}$ $c o m p a r i n g p o w e r s$ $2 π i = 0$ $π = 0 o r i = 0 ? ? ?$
Commented by abdo imad last updated on 17/Apr/18

$l o o k s i r t h e f o n c t i o n R \to] 0, + \infty [/ f (x) = e^{x} i s a b i j e c t i o n$ $f r o m R t o] 0, + \infty [f o r t h a t w e h a v e e^{x} = e^{y} \Leftrightarrow x = y$ $b u t t h e f u n c t i o n R \to C / g (x) = e^{i x} i s n o t a b i j e c t i o n a n d$ $e^{i x} = e^{i y} \Leftrightarrow x = y + 2 k π w i t h k \in Z a n d i f z = r e^{i θ}, r > 0$ $l n (z) = l n r + i (θ + 2 k π) b u t w e t a k e o n l y$ $l n (z) = l n r + i θ a s a p r i n c i p a l d e t e r m i n a t i o n o f t h e$ $c o m p l e x l o g a r i t h m e .$
Commented by Joel578 last updated on 17/Apr/18

$Same like \sin π = \sin 0$ $then π = 0 ? ?$ $Or {(- 1)}^{2} = 1^{2}$ $then - 1 = 1 ? ?$ $Guess why {it}^{'} s incorrect$
Commented by 33 last updated on 17/Apr/18

$b u t s i r,$ ${(- 1)}^{2} = 1^{2} ⇏ - 1 = 1$ $y o u a r e t a k i n g s q r o o t h e r e$ $w h i c h i s n o t v a l i d .$ $i d i d n o t t a k e s q u a r e r o o t .$ $A n d s i n e i s a$ $m a n y o n e f u n c t i o n .$ $f (x) = f (x + a)$ $⇏ x = x + a$ $f o r m a n y o n e f u n c t i o n s .$ $b u t e^{x} i s o n e o n e .$ $h e n c e e^{x} = e^{y} m u s t i m p l y$ $x = y$
	Answered by MJS last updated on 17/Apr/18

	$you {can}^{'} t compare the powers$ $in this case, because e^{φ i} is a$ $periodic function$ $z = a + b i = r (\cos φ + i \sin φ) = {r e}^{φ i}$ $[with r = abs (z) = \sqrt{a^{2} + b^{2}}$ $and φ = \tan^{- 1} \frac{b}{a}]$ $because \sin φ and \cos φ are periodic$ $it should be clear that for n \in N_{0}$ $z = r (\cos (φ \pm 2 n π) + i \sin (φ \pm 2 n π)) \Rightarrow$ $\Rightarrow z = {r e}^{φ \pm 2 n π}$ $therefore, e^{0} = e^{2 π i} ⇏ 0 = 2 π i$ $but you made another mistake :$ $x^{y} = - 1 ∣^{2}$ $x^{2 y} = 1$ $x^{2 y} = x^{0}$ $2 y = 0$ $y = 0$ $BUT x^{0} \neq - 1$ $\Rightarrow squaring both sides might$ $lead to wrong results$
	Commented by 33 last updated on 17/Apr/18

	$h a h a v e r y a p t e x p l a n a t i o n$ $s i r . t h a n k s f o r y o u r$ $e f f o r t .$
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