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

1 Answer
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):}