Find-the-value-of-1-3-3-4-help-please-

Question Number 85261 by Umar last updated on 20/Mar/20

Find the value of {(1 + \sqrt{- 3})}^{\frac{3}{4}}

help please

Commented by mathmax by abdo last updated on 20/Mar/20

A = {(1 + \sqrt{- 3})}^{\frac{3}{4}} = {(1 + i \sqrt{3})}^{\frac{3}{4}} b u t 1 + i \sqrt{3} = 2 (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2 e^{\frac{i π}{3}} \Rightarrow

A = {(2 e^{\frac{i π}{3}})}^{\frac{3}{4}} = 2^{\frac{3}{4}} e^{\frac{i π}{4}} =^{4} \sqrt{8} (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})

Answered by MJS last updated on 20/Mar/20

1 + \sqrt{- 3} = 1 + \sqrt{3} i = {2 e}^{i \frac{π}{3}}

{({2 e}^{i \frac{π}{3}})}^{\frac{3}{4}} = 2^{\frac{3}{4}} e^{i \frac{π}{4}} = 2^{\frac{1}{4}} + 2^{\frac{1}{4}} i

Commented by Umar last updated on 20/Mar/20

Thank you sir

Answered by Rio Michael last updated on 20/Mar/20

{(1 + \sqrt{- 3})}^{\frac{3}{4}} = {(1 + \sqrt{3} i)}^{\frac{3}{4}}

= 2 {(\cos (\frac{π}{3}) + isin (\frac{π}{3}))}^{\frac{3}{4}}

By demoives theorem, 2^{\frac{3}{4}} (\cos (\frac{π}{3} \times \frac{3}{4}) + isin (\frac{π}{3} \times \frac{3}{4})) = \sqrt[4]{8} (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)

these can be simplified and the answer gotton .

OR

you could expand

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots

\Rightarrow {(1 + \sqrt{- 3})}^{\frac{3}{4}} = 1 + \frac{3 \sqrt{- 3}}{4} + \frac{\frac{3}{4} (\frac{- 1}{4})}{2!} {(\sqrt{- 3})}^{2} + \dots

= 1 + \frac{3 \sqrt{- 3}}{4} + \frac{9}{32} + \dots

Commented by Umar last updated on 20/Mar/20

Thank you sir

Facebook Tweet Pin