solve for z∈C (z)^(1/2) =−1 (z)^(1/3) =−1 (z)^(1/4) =−1


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Question Number 61591 by MJS last updated on 05/Jun/19

$solve for z \in C$ $\sqrt[2]{z} = - 1$ $\sqrt[3]{z} = - 1$ $\sqrt[4]{z} = - 1$
Commented by Smail last updated on 05/Jun/19

$s a w \sqrt[2]{z}, \sqrt[3]{z} = - 1 a n d I t h o u g h t i t i s \sqrt[3]{(- 1)} a n d$ $s i n c e I w a s s o l v i n g t h i s p r o b l e m i n a r u s h$ $I m a d e t h a t s i l l y m i s t a k e .$ $M y a p o l o g i e s .$
Commented by maxmathsup by imad last updated on 05/Jun/19

$1) l e t z = {r e}^{i θ} \Rightarrow \sqrt{z} = \sqrt{r} e^{\frac{i θ}{2}} a n d \sqrt{z} = - 1 \Leftrightarrow \sqrt{z} = e^{i (2 k + 1) π} \Rightarrow \sqrt{r} e^{i \frac{θ}{2}} = e^{i (2 k + 1) π} \Rightarrow$ $r = 1 a n d \frac{θ}{2} = (2 k + 1) π \Rightarrow r = 1 a n d θ = (4 k + 2) π k \in Z \Rightarrow t h e s o l u t i o n a r e$ $z_{k} = e^{i (4 k + 2) π} k f r o m Z .$
Commented by Mr X pcx last updated on 05/Jun/19

$s i r s m a i l z^{\frac{1}{n}} = - 1 \Rightarrow z = {(- 1)}^{n}$ $i f z = e^{i π \frac{2 k + 1}{n}} \Rightarrow$ $z^{\frac{1}{n}} = e^{i π (\frac{2 k + 1}{n^{2}})} \neq - 1 s o y o u h a v e$ $c o m m i t e d a e r r o r . . .$
Commented by MJS last updated on 05/Jun/19
$this seems strange to me... obviously this is right: z^n =a with a∈R^− , n∈N^★ z=∣a∣^(1/n) e^(i((−π+2πk)/n)) with k∈N∧k<n z^3 =−1 ⇒ z=−1∨z=(1/2)±((√3)/2)i testing: (−1)^3 =−1; ((1/2)±((√3)/2)i)^2 =−1 now this z^(1/n) =a with a∈R^− , n∈N^★ should be similar: z=∣a∣^n e^(i(−π+2πk)n) with k∈N∧k<(1/n) ⇒ ⇒ k=0 ⇒ z=∣a∣^n e^(−iπn) = { ((−∣a∣^n with n=2m−1)),((∣a∣^n with n=2m)) :}; m∈N^★ z^(1/3) =−1 ⇒ z=−1 testing: ((−1))^(1/3) = { ((−1)),(((1/2)±((√3)/2)i)) :}; so it′s (1/3) right and (2/3) wrong and what about these: z^(3/2) =−1 z^(5/3) =−1 z^(√2) =−1$
$this seems strange to me . . .$ $obviously this is right :$ $z^{n} = a with a \in R^{-}, n \in N^{★}$ $z =∣ a ∣^{\frac{1}{n}} e^{i \frac{- π + 2 π k}{n}} with k \in N \land k < n$ $z^{3} = - 1 \Rightarrow z = - 1 \lor z = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $testing : {(- 1)}^{3} = - 1; {(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i)}^{2} = - 1$ $now this$ $z^{\frac{1}{n}} = a with a \in R^{-}, n \in N^{★}$ $should be similar :$ $z =∣ a ∣^{n} e^{i (- π + 2 π k) n} with k \in N \land k < \frac{1}{n} \Rightarrow$ $\Rightarrow k = 0 \Rightarrow z =∣ a ∣^{n} e^{- i π n} = {\begin{cases} - ∣ a ∣^{n} with n = 2 m - 1 \\ ∣ a ∣^{n} with n = 2 m \end{cases}; m \in N^{★}$ $z^{\frac{1}{3}} = - 1 \Rightarrow z = - 1$ $testing : \sqrt[3]{- 1} = {\begin{cases} - 1 \\ \frac{1}{2} \pm \frac{\sqrt{3}}{2} i \end{cases}; so {it}^{'} s \frac{1}{3} right and \frac{2}{3} wrong$ $and what about these :$ $z^{\frac{3}{2}} = - 1$ $z^{\frac{5}{3}} = - 1$ $z^{\sqrt{2}} = - 1$
Commented by Mr X pcx last updated on 05/Jun/19

$h e n e r a l l y l e t s o l v e^{n} \sqrt{z} = - 1$ $l e t z = {r e}^{i θ}$ $z^{\frac{1}{n}} = - 1 \Rightarrow r^{\frac{1}{n}} e^{i \frac{θ}{n}} = e^{i (2 k + 1) π} \Rightarrow$ $r = 1 a n d \frac{θ}{n} = (2 k + 1) π \Rightarrow$ $θ_{k} = (2 k + 1) n π \Rightarrow$ $z_{k} = e^{i (2 k + 1) n π} w i t h k f r o m Z .$ $b u t w e s e e t h a t z_{k} = e^{i (2 n k) π} e^{i n π} \Rightarrow$ $z_{k} = {(- 1)}^{n} . . . .!$
Commented by Mr X pcx last updated on 05/Jun/19

$g e n e r a l l y . . . .$
Commented by MJS last updated on 05/Jun/19

$I have serious doubts . it seems a matter of$ $definition . if {we}^{'} re not careful we might end$ $up with \sqrt{2} + \sqrt{2} = 0, not to mention what might$ $happen here : \sqrt[3]{- 1} + \sqrt[3]{- 1} + \sqrt[3]{- 1} = ? ? ?$
Commented by Smail last updated on 05/Jun/19

$I a m d e l e t i n g m y a n s w e r r i g h t n o w .$
	Answered by MJS last updated on 05/Jun/19

	Commented by MJS last updated on 05/Jun/19

	$if this is right (I believe it is) \Rightarrow {there}^{'} s no$ $n \in Z for some cases \Rightarrow these cases have no$ $solution$ $z^{\frac{1}{k}} = - 1 with z \in C \land k \in N^{★}$ $\Rightarrow \frac{1}{2} - \frac{1}{2 k} ⩽ n < \frac{1}{2} + \frac{1}{2 k}$ $this only has a solution for k = 1 \Rightarrow 0 ⩽ n < 1 \Rightarrow n = 0$
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