solve-for-z-C-z-1-2-1-z-1-3-1-z-1-4-1-

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 \dots

Commented by MJS last updated on 05/Jun/19

this seems strange to me \dots

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} \dots .!

Commented by Mr X pcx last updated on 05/Jun/19

g e n e r a l l y \dots .

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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