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Question Number 211383 by zakirullah last updated on 07/Sep/24

Commented by zakirullah last updated on 07/Sep/24

$p l e a s e n e e d d i f f e r e n c e w i t h s o l u t i o n .$
	Answered by A5T last updated on 07/Sep/24

	$\sqrt{{(- 9)}^{2}} = \sqrt{- 9 \times - 9} = \sqrt{81} = 9 ✓$ $\sqrt{- 9 \times - 9} \neq \sqrt{- 9} \times \sqrt{- 9} = {(\sqrt{- 9})}^{2}$ ${(\sqrt{- 9})}^{2} = {(3 i)}^{2} = - 9 ✓$ ${(\sqrt{- 9})}^{2} = \sqrt{- 9} \times \sqrt{- 9} \neq \sqrt{- 9 \times - 9}$ $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b} w h e n a a n d b a r e n o t b o t h < 0$
	Commented by zakirullah last updated on 07/Sep/24

	$a b o u n d l e o f t h a n k s$
	Answered by Frix last updated on 07/Sep/24

	$- 9 = {9 e}^{i π}$ $\sqrt{{(- 9)}^{2}} = \sqrt{{({9 e}^{i π})}^{2}} = \sqrt{81 \underset{()}{\underset{⏟}{e^{2 i π}}}} = \sqrt{{81 e}^{0}} = \sqrt{81} = 9$ ${(\sqrt{- 9})}^{2} = {(\sqrt{{9 e}^{i π}})}^{2} = {({3 e}^{i \frac{π}{2}})}^{2} = {9 e}^{i π} = - 9$ $() e^{2 i π} is just an intermediate step result .$ $Before the next calculation we have to make$ $sure the angle is \in (- π, π] .$ $This rule lately got lost on its path through$ $the intellectual night of the www . . .$ $z \in C : z = r e^{i θ} \land r ⩾ 0 \land - π < θ ⩽ π$ $\Rightarrow sometimes {(z^{q})}^{\frac{1}{q}} \neq z$ $Example :$ ${(1 + i)}^{5} = {(\sqrt{2} e^{i \frac{π}{4}})}^{5} = 4 \sqrt{2} e^{i \frac{5 π}{4}} = 4 \sqrt{2} e^{- i \frac{3 π}{4}} =$ $= - 4 - 4 i$ ${(- 4 - 4 i)}^{\frac{1}{5}} = {(4 \sqrt{2} e^{- \frac{3 π}{4}})}^{\frac{1}{5}} = \sqrt{2} e^{- i \frac{3 π}{20}} =$ $= \sqrt{2} (\cos \frac{3 π}{20} - \sin \frac{3 π}{20}) \approx 1.26 - . 643 i$
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