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Question Number 110608 by Engr_Jidda last updated on 29/Aug/20

\sqrt{x} + 1 = 0

S o l u t i o n

\sqrt{x} = - 1

r e c a l l e d t h a t ϱ^{i Π} = - 1

\sqrt{x} = ϱ^{i Π}

x = {(ϱ^{i Π})}^{2}

\Rightarrow x = ϱ^{2 i Π}

o r x = 2 c o s Π + i s i n Π

x h a s n o r e a l v a l u e!

Commented by Her_Majesty last updated on 29/Aug/20

{y o u}^{'} r e w r o n g

x = e^{2 i π} \Leftrightarrow x = c o s 2 π + i s i n 2 π = 1

a n y w a y {y o u}^{'} r e w r o n g b e c a u s e w e a r e n o t

a l l o w e d t o s q u a r e b o t h s i d e s o f a n y e q u a t i o n;

t h i s l e a d s t o f a l s e s o l u t i o n s :

x = 3

x^{2} = 9 \Rightarrow x = \pm 3

i n o u r c a s e w e h a v e t o s e t r a n g e s f i r s t

i f x \in C \Rightarrow x = {r e}^{i θ} w i t h r \in R \land r ⩾ 0 a n d

θ \in R \land {\begin{cases} - π ⩽ θ < π \\ 0 ⩽ θ < 2 π \end{cases} (b o t h a r e i n u s e)

\sqrt{x} = - 1

\sqrt{{r e}^{i θ}} = - 1

\sqrt{r} e^{i θ / 2} = - 1

b u t - 1 = e^{\pm i π}

\sqrt{r} e^{i θ / 2} = e^{\pm i π} \Leftrightarrow r = 1 \land θ / 2 = \pm π \Leftrightarrow θ = \pm 2 π

b u t θ i s n o t i n o u r r a n g e \Rightarrow w e h a v e t o

s u b t r a c t / a d d 2 π \Rightarrow θ = 0

\Rightarrow x = \sqrt{x} = e^{0} = 1 \neq - 1 \Rightarrow n o s o l u t i o n a t a l l