Let
a=5-i
IaI=52+(−1)2\sqrt{ {5}^2+({-1})^2}\\52+(−1)2
IaI=26\sqrt{26}\\26
α=tan−1(−5)a=∣a∣eiα=∣a∣(cosα+isinα)a15=∣a∣15eiα5a15=2615(costan−1(1−5)+2kπ5+isintan−1(1−5)+2kπ5)k=0,1,2,3,4\alpha = tan^{-1}(-5)\\ a=|a|e^{i\alpha}=|a|(cos\alpha+isin\alpha)\\ a^{\frac{1}{5}}=|a|^{\frac{1}{5}}e^{i\frac{\alpha}{5}}\\ a^{\frac{1}{5}}=\sqrt{26}^{\frac{1}{5}}(cos\frac{tan^{-1}(\frac{1}{-5})+2kπ}{5}+isin\frac{tan^{-1}(\frac{1}{-5})+2kπ}{5})\\ k= 0, 1, 2, 3, 4α=tan−1(−5)a=∣a∣eiα=∣a∣(cosα+isinα)a51=∣a∣51ei5αa51=2651(cos5tan−1(−51)+2kπ+isin5tan−1(−51)+2kπ)k=0,1,2,3,4
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment