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Question Number 120679 by kaivan.ahmadi last updated on 02/Nov/20

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Commented by bobhans last updated on 01/Nov/20

\sqrt[3]{- 8}

(1.0 + 1 . 732051 i)

Commented by bobhans last updated on 01/Nov/20

the result from calculator

Commented by mathmax by abdo last updated on 02/Nov/20

if a > 0 and a \neq 1 we have a^{x} = e^{xln (a)}

if a < o \Rightarrow a^{x} = {(- (- a))}^{x} = {(- 1)}^{x} {(- a)}^{x} = {(e^{i π})}^{x} e^{xln (- a)}

= e^{i π x} e^{xln (- a)} = (\cos π x + isin (π x) e^{xln (- a)} (- a > 0)

or a^{x} = e^{xln (- a) + i π x}

Answered by MJS_new last updated on 01/Nov/20

but {(- 8)}^{\frac{1}{3}} i s defined .

there are 2 definitions

(1) z \in C, r \in R^{+} : z^{\frac{1}{n}} = {(r e^{i θ})}^{\frac{1}{n}} = r^{\frac{1}{n}} e^{i \frac{θ}{n}}

(2) for θ = π \land n = 2 k + 1 we have z = r e^{i π} = - r

and define z^{\frac{1}{2 k + 1}} = {(- r)}^{\frac{1}{2 k + 1}} = - (r^{\frac{1}{2 k + 1}})

this 2^{nd} definition is “ {older}^{″} because in

basic mathematics {it}^{'} s common to use it

\Rightarrow {(- 8)}^{\frac{1}{3}} = {\begin{cases} 1 + \sqrt{3} i \\ - 2 \end{cases}

Commented by bobhans last updated on 01/Nov/20

but sir why if we use a calculator

give the answer 1 + i \sqrt{3} ?

Commented by MJS_new last updated on 02/Nov/20

depends on the calculator . the ones for

high schools should give - 2, the ones for

university give 1 + \sqrt{3} i

I guess using the exceptions \sqrt[3]{- 27} = - 3;

\sqrt[5]{- 32} = - 2 e t c . makes it impossible to

get a result for calculations like {(- 1)}^{\frac{2}{3}}

and the exeptions do harm to functions

like f (x) : y = {(- 1)}^{x} for x \in R \land y \in C

\Rightarrow most advanced calculators use

z^{p + q i} = {(r e^{i θ})}^{p + q i} = r^{p + q i} e^{- q θ + ip θ} =

= \frac{r^{p}}{e^{q θ}} e^{i (p θ + q \ln r)} which for q = 0 gives r^{p} e^{i p θ}

Commented by bobhans last updated on 02/Nov/20

thank you sir

Answered by malwaan last updated on 02/Nov/20

{(- 8)}^{\frac{1}{3}} i s d e f i n e d

x = {(- 8)}^{\frac{1}{3}}

\Rightarrow x^{3} = - 8 \Rightarrow x^{3} + 8 = 0

(x + 2) (x^{2} - 2 x + 4) = 0

\Rightarrow x = - 2

a n d x = 1 \pm \sqrt{3} i

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