RealiseSomethin
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- 2013
lol hard root..
its usually very unclean... for instance, in the example (12 - 6i) the value of the argument is -tan^(-1) [2], which you'd be struggling to find any sort of nice form for - you can plug it in your calculator, halve it and find the cartesian form of the square root but either you're going to have an answer in terms of tan^-1 stuff, or one in the decimal form your calculator puts it in, neither of which I think are usually accepted as answers. You could do it by solving to find sin(theta/2) and cos(theta/2) from the quadratic equations you get from the double angle formulae, but thats effectively the same thing as doing it the way slyhunter didIsn't it just plus or minus the square root of the modulus times cis(theta) where theta is half the argument? That seems like a much more simplistic way considering the argument is so clean.
No.... Tan@=y/x so it's -0.5Isn't the value of the argument -pi/4? That isn't difficult. The Square root is plus or minus the fourth root of 180 times cis -pi/8. Relatively simple considering the alternative solution given.
No, tan@=-0.5, therefore @=pi/4. When you find the square root, you divide the argument by 2, Hence @=-pi/8.No.... Tan@=y/x so it's -0.5
Erm...how the hell does = ? ..... It doesn't; check with your calculator.No, tan@=-0.5, therefore @=pi/4. When you find the square root, you divide the argument by 2, Hence @=-pi/8.
From memory, I think it's a Patel question. A few answers are just the argument left as an arctan, and not really give you a proper answer.From where did you get this question? I'm curious as to what kind of textbook would give you a question with a solution so ugly.
My bad, got confused with sin(45). Im an idiot.Erm...how the hell does = ? ..... It doesn't; check with your calculator.
Its a bad dyslexia day.lol either way, sin45 = 1/sqrt2, not -0.5
LMAAOOOOOIts a bad dyslexia day.