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

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

$$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\sqrt[{\mathrm{2}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{z}}=−\mathrm{1} \\ $$

Commented by Smail last updated on 05/Jun/19

saw (z)^(1/2) ,(z)^(1/3) =−1 and I thought it is (((−1)))^(1/3) and since Iwas solving this problem in a rush I made that silly mistake. My apologies.

$$\:{saw}\:\sqrt[{\mathrm{2}}]{{z}}\:,\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1}\:{and}\:{I}\:{thought}\:{it}\:{is}\:\sqrt[{\mathrm{3}}]{\left(−\mathrm{1}\right)}\:\:\:{and}\: \\ $$$${since}\:{Iwas}\:{solving}\:{this}\:{problem}\:{in}\:{a}\:{rush} \\ $$$${I}\:{made}\:{that}\:{silly}\:{mistake}. \\ $$$${My}\:{apologies}. \\ $$

Commented by maxmathsup by imad last updated on 05/Jun/19

1) let z =re^(iθ) ⇒(√z)=(√r)e^((iθ)/2) and (√z)=−1 ⇔(√z)=e^(i(2k+1)π) ⇒(√r)e^(i(θ/2)) =e^(i(2k+1)π) ⇒ r=1 and (θ/2) =(2k+1)π ⇒r=1 and θ =(4k+2)π k ∈ Z ⇒the solution are z_k =e^(i(4k+2)π) k from Z .

$$\left.\mathrm{1}\right)\:{let}\:{z}\:={re}^{{i}\theta} \:\Rightarrow\sqrt{{z}}=\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} \:\:\:{and}\:\sqrt{{z}}=−\mathrm{1}\:\Leftrightarrow\sqrt{{z}}={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow\sqrt{{r}}{e}^{{i}\frac{\theta}{\mathrm{2}}} \:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow \\ $$$${r}=\mathrm{1}\:\:{and}\:\frac{\theta}{\mathrm{2}}\:=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:\:\:\:\:\Rightarrow{r}=\mathrm{1}\:{and}\:\theta\:=\left(\mathrm{4}{k}+\mathrm{2}\right)\pi\:\:\:\:{k}\:\in\:{Z}\:\:\Rightarrow{the}\:{solution}\:{are} \\ $$$${z}_{{k}} ={e}^{{i}\left(\mathrm{4}{k}+\mathrm{2}\right)\pi} \:\:\:\:\:{k}\:{from}\:{Z}\:. \\ $$

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

sir smail z^(1/n) =−1 ⇒z =(−1)^n if z=e^(iπ((2k+1)/n)) ⇒ z^(1/n) =e^(iπ(((2k+1)/n^2 ))) ≠−1 so you have commited a error...

$${sir}\:{smail}\:\:{z}^{\frac{\mathrm{1}}{{n}}} =−\mathrm{1}\:\Rightarrow{z}\:=\left(−\mathrm{1}\right)^{{n}} \\ $$$${if}\:{z}={e}^{{i}\pi\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}} \:\:\:\Rightarrow \\ $$$${z}^{\frac{\mathrm{1}}{{n}}} ={e}^{{i}\pi\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}^{\mathrm{2}} }\right)} \:\neq−\mathrm{1}\:{so}\:{you}\:{have} \\ $$$${commited}\:{a}\:{error}... \\ $$

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

$$\mathrm{this}\:\mathrm{seems}\:\mathrm{strange}\:\mathrm{to}\:\mathrm{me}... \\ $$$$ \\ $$$$\mathrm{obviously}\:\mathrm{this}\:\mathrm{is}\:\mathrm{right}: \\ $$$${z}^{{n}} ={a}\:\mathrm{with}\:{a}\in\mathbb{R}^{−} ,\:{n}\in\mathbb{N}^{\bigstar} \\ $$$${z}=\mid{a}\mid^{\frac{\mathrm{1}}{{n}}} \mathrm{e}^{\mathrm{i}\frac{−\pi+\mathrm{2}\pi{k}}{{n}}} \:\mathrm{with}\:{k}\in\mathbb{N}\wedge{k}<{n} \\ $$$$ \\ $$$${z}^{\mathrm{3}} =−\mathrm{1}\:\Rightarrow\:{z}=−\mathrm{1}\vee{z}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{testing}:\:\left(−\mathrm{1}\right)^{\mathrm{3}} =−\mathrm{1};\:\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{this} \\ $$$${z}^{\frac{\mathrm{1}}{{n}}} ={a}\:\mathrm{with}\:{a}\in\mathbb{R}^{−} ,\:{n}\in\mathbb{N}^{\bigstar} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{similar}: \\ $$$${z}=\mid{a}\mid^{{n}} \mathrm{e}^{\mathrm{i}\left(−\pi+\mathrm{2}\pi{k}\right){n}} \:\mathrm{with}\:{k}\in\mathbb{N}\wedge{k}<\frac{\mathrm{1}}{{n}}\:\Rightarrow \\ $$$$\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{z}=\mid{a}\mid^{{n}} \mathrm{e}^{−\mathrm{i}\pi{n}} =\begin{cases}{−\mid{a}\mid^{{n}} \:\mathrm{with}\:{n}=\mathrm{2}{m}−\mathrm{1}}\\{\mid{a}\mid^{{n}} \:\mathrm{with}\:{n}=\mathrm{2}{m}}\end{cases};\:{m}\in\mathbb{N}^{\bigstar} \\ $$$$ \\ $$$${z}^{\frac{\mathrm{1}}{\mathrm{3}}} =−\mathrm{1}\:\Rightarrow\:{z}=−\mathrm{1} \\ $$$$\mathrm{testing}:\:\sqrt[{\mathrm{3}}]{−\mathrm{1}}=\begin{cases}{−\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}}\end{cases};\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{right}\:\mathrm{and}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{wrong} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{what}\:\mathrm{about}\:\mathrm{these}: \\ $$$${z}^{\frac{\mathrm{3}}{\mathrm{2}}} =−\mathrm{1} \\ $$$${z}^{\frac{\mathrm{5}}{\mathrm{3}}} =−\mathrm{1} \\ $$$${z}^{\sqrt{\mathrm{2}}} =−\mathrm{1} \\ $$

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

henerally let solve^n (√z)=−1 let z =re^(iθ) z^(1/n) =−1 ⇒r^(1/n) e^(i(θ/n)) =e^(i(2k+1)π) ⇒ r=1 and (θ/n) =(2k+1)π ⇒ θ_k =(2k+1)nπ ⇒ z_k =e^(i(2k+1)nπ) with k from Z . but we see that z_k =e^(i(2nk)π) e^(inπ) ⇒ z_k =(−1)^n ....!

$${henerally}\:{let}\:{solve}\:^{{n}} \sqrt{{z}}=−\mathrm{1} \\ $$$${let}\:{z}\:={re}^{{i}\theta} \:\: \\ $$$${z}^{\frac{\mathrm{1}}{{n}}} =−\mathrm{1}\:\Rightarrow{r}^{\frac{\mathrm{1}}{{n}}} \:{e}^{{i}\frac{\theta}{{n}}} \:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow \\ $$$${r}=\mathrm{1}\:{and}\:\:\frac{\theta}{{n}}\:=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\Rightarrow \\ $$$$\theta_{{k}} =\left(\mathrm{2}{k}+\mathrm{1}\right){n}\pi\:\Rightarrow \\ $$$${z}_{{k}} ={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right){n}\pi} \:\:\:\:{with}\:{k}\:{from}\:{Z}\:. \\ $$$${but}\:{we}\:{see}\:{that}\:{z}_{{k}} ={e}^{{i}\left(\mathrm{2}{nk}\right)\pi} \:{e}^{{in}\pi} \:\Rightarrow \\ $$$${z}_{{k}} =\left(−\mathrm{1}\right)^{{n}} \:....! \\ $$

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

generally....

$${generally}.... \\ $$

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 (√2)+(√2)=0, not to mention what might happen here: ((−1))^(1/3) +((−1))^(1/3) +((−1))^(1/3) =???

$$\mathrm{I}\:\mathrm{have}\:\mathrm{serious}\:\mathrm{doubts}.\:\mathrm{it}\:\mathrm{seems}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of} \\ $$$$\mathrm{definition}.\:\mathrm{if}\:\mathrm{we}'\mathrm{re}\:\mathrm{not}\:\mathrm{careful}\:\mathrm{we}\:\mathrm{might}\:\mathrm{end} \\ $$$$\mathrm{up}\:\mathrm{with}\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}=\mathrm{0},\:\mathrm{not}\:\mathrm{to}\:\mathrm{mention}\:\mathrm{what}\:\mathrm{might} \\ $$$$\mathrm{happen}\:\mathrm{here}:\:\sqrt[{\mathrm{3}}]{−\mathrm{1}}+\sqrt[{\mathrm{3}}]{−\mathrm{1}}+\sqrt[{\mathrm{3}}]{−\mathrm{1}}=??? \\ $$

Commented by Smail last updated on 05/Jun/19

I am deleting my answer right now.

$${I}\:{am}\:{deleting}\:{my}\:{answer}\:{right}\:{now}. \\ $$

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) ⇒ there′s no n∈Z for some cases ⇒ these cases have no solution z^(1/k) =−1 with z∈C∧k∈N^★ ⇒ (1/2)−(1/(2k))≤n<(1/2)+(1/(2k)) this only has a solution for k=1 ⇒ 0≤n<1 ⇒ n=0

$$\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{right}\:\left(\mathrm{I}\:\mathrm{believe}\:\mathrm{it}\:\mathrm{is}\right)\:\Rightarrow\:\mathrm{there}'\mathrm{s}\:\mathrm{no} \\ $$$${n}\in\mathbb{Z}\:\mathrm{for}\:\mathrm{some}\:\mathrm{cases}\:\Rightarrow\:\mathrm{these}\:\mathrm{cases}\:\mathrm{have}\:\mathrm{no} \\ $$$$\mathrm{solution} \\ $$$${z}^{\frac{\mathrm{1}}{{k}}} =−\mathrm{1}\:\mathrm{with}\:{z}\in\mathbb{C}\wedge{k}\in\mathbb{N}^{\bigstar} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{k}}\leqslant{n}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$$\mathrm{this}\:\mathrm{only}\:\mathrm{has}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for}\:{k}=\mathrm{1}\:\Rightarrow\:\mathrm{0}\leqslant{n}<\mathrm{1}\:\Rightarrow\:{n}=\mathrm{0} \\ $$

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