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Hard Root of Complex Number (1 Viewer)

largarithmic

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Isn'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.
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 did
 

lolcakes52

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Isn'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.
 

D94

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Isn'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@=y/x so it's -0.5
 

D94

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No, tan@=-0.5, therefore @=pi/4. When you find the square root, you divide the argument by 2, Hence @=-pi/8.
Erm...how the hell does = ? ..... It doesn't; check with your calculator.
 

Carrotsticks

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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.
 

D94

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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.
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.
 

Nooblet94

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Do you realise you just replied to a post that's 2.5 months old..?
 

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