write-1-i-23-3-i-13-in-re-i-

Question Number 132950 by mohammad17 last updated on 17/Feb/21

w r i t e \frac{{(1 - i)}^{23}}{{(\sqrt{3} - i)}^{13}} i n ({r e}^{i θ})

Answered by metamorfose last updated on 17/Feb/21

z = \frac{{(1 - i)}^{23}}{{(\sqrt{3} - i)}^{13}} = \frac{1}{2 \sqrt{2}} e^{i \frac{5 π}{12}}

Commented by metamorfose last updated on 17/Feb/21

o k . .

z = \frac{{(1 - i)}^{23}}{{(\sqrt{3} - i)}^{13}} = \frac{{(\sqrt{2} e^{- i \frac{π}{4}})}^{23}}{{(2 e^{- i \frac{π}{6}})}^{13}} = \frac{{(\sqrt{2})}^{23} e^{- i \frac{23 π}{4}}}{2^{13} e^{- i \frac{13 π}{6}}}

= \frac{{(\sqrt{2})}^{23}}{{(\sqrt{2})}^{26}} e^{- i (\frac{23 π}{4} - \frac{13 π}{6})} = \frac{1}{{(\sqrt{2})}^{3}} e^{- i \frac{π}{2} (\frac{23}{2} - \frac{13}{3})}

= \frac{1}{2 \sqrt{2}} e^{- i \frac{43 π}{12}} = \frac{1}{2 \sqrt{2}} e^{- i (\frac{48 π}{12} - \frac{5 π}{12})} = \frac{1}{2 \sqrt{2}} e^{- i (4 π - \frac{5 π}{12})}

= \frac{1}{2 \sqrt{2}} e^{i \frac{5 π}{12}}

Commented by mohammad17 last updated on 17/Feb/21

s i r c a n y o u g i v e m e s t e b s

Commented by mohammad17 last updated on 17/Feb/21

i w a n t s t e b b y s t e b

Commented by mohammad17 last updated on 18/Feb/21

s i r w h e r e e^{- 4 π i}

Commented by metamorfose last updated on 18/Feb/21

e^{- 4 i π} = 1