SOLUTION: The square root of -16/(3-3i)+(1-2i) I know that the square rot of -16 is 4i but Idk what to do from there. PLEASE HELP!!

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: The square root of -16/(3-3i)+(1-2i) I know that the square rot of -16 is 4i but Idk what to do from there. PLEASE HELP!!      Log On


   

Question 831919: The square root of -16/(3-3i)+(1-2i)
I know that the square rot of -16 is 4i but Idk what to do from there. PLEASE HELP!!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The square root of -16/(3-3i)+(1-2i)
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[4i/(3-3i) + 1-2i
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= [(4i(3+3i)/(9+9)] + 1-2i
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= [4i*3(1+i)/18] + 1-2i
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= (2/3)(1+i) + 1 - 2i
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= (5/3)-(4/3)i
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Cheers,
Stan H.
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