Answer to Question #2824 in Abstract Algebra for alyssa

Question

Square root of 11 + 5i in polar form with r greater than or equal to 0 and theta in the interval 0 less than or equal to theta less than or equal to 2 pie and rounded to the nearest thousandth.

Answer 1

Suppose
√(11+5i)=r(cos +isinφ)
We have that
11+5i=A(cosφ+isinφ), where
A = |11+5| =√(11²+5² )=√(121+25)=√146 = 12.08

φ = arg ( 11 + 5i) = arctan(5/11) = 0.427

Then
√(11+5i) = √A (cos(φ/2)+isin(φ/2))

√A=√12.083= 3.476
φ/2=0.213
cos(φ/2)=0.977
sin(φ/2)=0.212

Hence
√(11+5i)=3.476 (0.977+i 0.212)= 3.397 + i 0.736