How do you find the 4rd root of #16e^(90i)#?

1 Answer
Alan P.
Mar 10, 2018

#root(4)(e^(90^circi)) in {+-[cos(22.5^circ)+i * sin(22.5^circ)],+-[i *cos(22.5^circ)-sin(22.5^circ)]}#

Explanation:

Assuming the #90# was meant to be in degrees:

#e^(90^circi) = cos(90^circ)+i * sin(90^circ)color(white)("xxxx")#[Euler's formula]

Using De Moivre's formula
#sqrt(cos(90^circ)+i * sin(90^circ))=+-[cos(45^circ)+i * sin(45^circ)]#

#root(4)(e^(90^circ)i)=sqrt(+-[cos(45^circ)+i * sin(45^circ)])#

Re-applying deMoivre's formula
#{: (sqrt(+[cos(45^circ)+i * sin(45^circ)]),color(white)("xx"),sqrt(-[cos(45^circ)+i * sin(45^circ)])), (=+-[cos(22.5^circ)+i * sin(22.5^circ)],,=+-i * [cos(22.5^circ)+i * sin(22.5^circ)]), (,,=+-[i *cos(22.5^circ)-sin(22.5^circ)]) :}#

For approximate values you could substitute the approximations:
#{:(cos(22.5^circ)~~0.9239,color(white)("xx")sin(22.5^circ)~~0.3827):}#

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