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.
1
Expert's answer
2011-05-19T07:12:19-0400
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))
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Leave a comment