Question Number 211383 by zakirullah last updated on 07/Sep/24
Commented by zakirullah last updated on 07/Sep/24
$${please}\:{need}\:{difference}\:{with}\:{solution}. \\ $$
Answered by A5T last updated on 07/Sep/24
$$\sqrt{\left(−\mathrm{9}\right)^{\mathrm{2}} }=\sqrt{−\mathrm{9}×−\mathrm{9}}=\sqrt{\mathrm{81}}=\mathrm{9}\:\:\:\:\checkmark \\ $$$$\sqrt{−\mathrm{9}×−\mathrm{9}}\neq\sqrt{−\mathrm{9}}×\sqrt{−\mathrm{9}}=\left(\sqrt{−\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\left(\sqrt{−\mathrm{9}}\right)^{\mathrm{2}} =\left(\mathrm{3}{i}\right)^{\mathrm{2}} =−\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\checkmark \\ $$$$\left(\sqrt{−\mathrm{9}}\right)^{\mathrm{2}} =\sqrt{−\mathrm{9}}×\sqrt{−\mathrm{9}}\neq\sqrt{−\mathrm{9}×−\mathrm{9}} \\ $$$$ \\ $$$$\sqrt{{a}×{b}}=\sqrt{{a}}×\sqrt{{b}}\:{when}\:{a}\:{and}\:{b}\:{are}\:{not}\:{both}\:<\mathrm{0} \\ $$
Commented by zakirullah last updated on 07/Sep/24
$${a}\:{boundle}\:{of}\:{thanks} \\ $$
Answered by Frix last updated on 07/Sep/24
![−9=9e^(iπ) (√((−9)^2 ))=(√((9e^(iπ) )^2 ))=(√(81e^(2iπ) _((∗)) ))=(√(81e^0 ))=(√(81))=9 ((√(−9)))^2 =((√(9e^(iπ) )))^2 =(3e^(i(π/2)) )^2 =9e^(iπ) =−9 (∗) e^(2iπ) is just an intermediate step result. Before the next calculation we have to make sure the angle is ∈(−π, π]. This rule lately got lost on its path through the intellectual night of the www... z∈C: z=re^(iθ) ∧r≥0∧−π<θ≤π ⇒ sometimes (z^q )^(1/q) ≠z Example: (1+i)^5 =((√2)e^(i(π/4)) )^5 =4(√2)e^(i((5π)/4)) =4(√2)e^(−i((3π)/4)) = =−4−4i (−4−4i)^(1/5) =(4(√2)e^(−((3π)/4)) )^(1/5) =(√2)e^(−i((3π)/(20))) = =(√2)(cos ((3π)/(20)) −sin ((3π)/(20)))≈1.26−.643i](https://www.tinkutara.com/question/Q211386.png)
$$−\mathrm{9}=\mathrm{9e}^{\mathrm{i}\pi} \\ $$$$\sqrt{\left(−\mathrm{9}\right)^{\mathrm{2}} }=\sqrt{\left(\mathrm{9e}^{\mathrm{i}\pi} \right)^{\mathrm{2}} }=\sqrt{\mathrm{81}\underset{\left(\ast\right)} {\underbrace{\mathrm{e}^{\mathrm{2i}\pi} }}}=\sqrt{\mathrm{81e}^{\mathrm{0}} }=\sqrt{\mathrm{81}}=\mathrm{9} \\ $$$$\left(\sqrt{−\mathrm{9}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{9e}^{\mathrm{i}\pi} }\right)^{\mathrm{2}} =\left(\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\mathrm{2}} =\mathrm{9e}^{\mathrm{i}\pi} =−\mathrm{9} \\ $$$$ \\ $$$$\left(\ast\right)\:\mathrm{e}^{\mathrm{2i}\pi} \:\mathrm{is}\:\mathrm{just}\:\mathrm{an}\:\mathrm{intermediate}\:\mathrm{step}\:\mathrm{result}. \\ $$$$\mathrm{Before}\:\mathrm{the}\:\mathrm{next}\:\mathrm{calculation}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{make} \\ $$$$\mathrm{sure}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{is}\:\in\left(−\pi,\:\pi\right]. \\ $$$$\mathrm{This}\:\mathrm{rule}\:\mathrm{lately}\:\mathrm{got}\:\mathrm{lost}\:\mathrm{on}\:\mathrm{its}\:\mathrm{path}\:\mathrm{through} \\ $$$$\mathrm{the}\:\mathrm{intellectual}\:\mathrm{night}\:\mathrm{of}\:\mathrm{the}\:\mathrm{www}… \\ $$$$ \\ $$$${z}\in\mathbb{C}:\:{z}={r}\mathrm{e}^{\mathrm{i}\theta} \wedge{r}\geqslant\mathrm{0}\wedge−\pi<\theta\leqslant\pi \\ $$$$\Rightarrow\:\mathrm{sometimes}\:\left({z}^{{q}} \right)^{\frac{\mathrm{1}}{{q}}} \neq{z} \\ $$$$\mathrm{Example}: \\ $$$$\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{5}} =\left(\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{5}} =\mathrm{4}\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} =\mathrm{4}\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} = \\ $$$$=−\mathrm{4}−\mathrm{4i} \\ $$$$\left(−\mathrm{4}−\mathrm{4i}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} =\left(\mathrm{4}\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{3}\pi}{\mathrm{4}}} \right)^{\frac{\mathrm{1}}{\mathrm{5}}} =\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{20}}} = \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{20}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{20}}\right)\approx\mathrm{1}.\mathrm{26}−.\mathrm{643i} \\ $$