if-a-1-i-3-find-a-1-5-

Question Number 170675 by MathsFan last updated on 29/May/22

if a = - 1 - i \sqrt{3}

find a^{\frac{1}{5}}

Answered by Mathspace last updated on 29/May/22

∣ a ∣= \sqrt{1 + 3} = 2 \Rightarrow a = 2 (- \frac{1}{2} - \frac{\sqrt{3}}{2} i)

= 2 e^{i (\frac{4 π}{3})} \Rightarrow a^{\frac{1}{5}} = 2^{\frac{1}{5}} e^{i \frac{4 π}{15}}

=^{5} \sqrt{2} {c o s (\frac{4 π}{15}) + i s i n (\frac{4 π}{15})}

Commented by MathsFan last updated on 29/May/22

t h a n k s

Answered by mr W last updated on 29/May/22

a = 2 [\cos (\frac{4 π}{3} + 2 k π) + \sin (\frac{4 π}{3} + 2 k π) i]

\Rightarrow a^{\frac{1}{5}} = \sqrt[5]{2} [\cos (\frac{4 π}{15} + \frac{2 k π}{5}) + \sin (\frac{4 π}{15} + \frac{2 k π}{5}) i]

(k = 0, 1, 2, 3, 4)

Commented by MJS_new last updated on 29/May/22

back when I studied mathematics this would

have been wrong . {it}^{'} s a difference between

the results of

(1) \sqrt{4}

(2) x^{2} = 4

because (1) is just a simple calculation but

(2) is an equation where we want all possible

values of x satisfying the eq .

(1) \sqrt{4} = 2 is unique and not \sqrt{4} = \pm 2

(2) x^{2} = 4 \Rightarrow x = \pm 2

same for z^{1 / 5} versus x^{5} = z with z = r e^{i θ}

z^{1 / 5} = r^{1 / 5} e^{i θ / 5} is unique

x^{5} = z has the obvious 5 solutions

Commented by mr W last updated on 30/May/22

b a s i c a l l y y o u a r e r i g h t s i r . b u t i c a s e

o f c o m p l e x n u m b e r s i t h i n k z^{\frac{1}{5}} m a y

h a v e i n d e e d 5 d i f f e r e n t v a l u e s .

z = {r e}^{i θ} = {r e}^{i (θ + 2 k π)} w i t h k \in Z

\Rightarrow z^{\frac{1}{5}} = \sqrt[5]{r} e^{i (\frac{θ}{5} + \frac{2 k π}{5})}

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