what is solution of (198+10√(-1))^(1/3) as form a+ib

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2 Answers

konvert 198+10i tu radial form . . . 198^2=39,204+10^2=100, then sqroot(39304)=198.2523644 . . . =leng av vekter . . . tan-1((10/198)=tan-1(0.0505050505)=2.891269596 deg . . . then (...)^1/3: kube root(198.2523644...)=5.830951884 . . . kube root=kut angel bi 3, so 2.891269596/3=0.963756532 deg . . . then, reel komponent=kosine(0.96... deg)*length . . . reel=5.830951884*09999858534=5.830127008 . . . imajinary part=5.83...* sine(0.96375...)=5.83...*0.016819931 . .. =0.098062336
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12556K-1245= 235789
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