Menu Close

4-6-i-5-4-polar-




Question Number 124474 by sogol last updated on 03/Dec/20
((4/(−6+i(√5))))^4 =???    polar????
(46+i5)4=???polar????
Commented by MJS_new last updated on 03/Dec/20
no need to convert to polar  (a+bi)^4 =a^4 −6a^2 b^2 +b^4 +4xy(x^2 −y^2 )i  (1/(a+bi))=((a−bi)/(a^2 +b^2 ))  ((4/(−6+i(√5))))^4 =((256)/(241−744(√5)i))=  =((61696)/(2825761))+((190464(√5))/(2825761))i  it′s not easier in polar
noneedtoconverttopolar(a+bi)4=a46a2b2+b4+4xy(x2y2)i1a+bi=abia2+b2(46+i5)4=2562417445i==616962825761+19046452825761iitsnoteasierinpolar
Answered by mathmax by abdo last updated on 03/Dec/20
we have  ∣−6+i(√5)∣=(√(36+5))=(√(41)) ⇒−6+i(√5)=(√(41)) e^(−iarctan(((√5)/6)))  ⇒  ((4/(−6+i(√5))))^4  =(4^4 /(((√(41)))^4  e^(−4iarctan(((√5)/6))) )) =(4^4 /(((√(41)))^4 )) e^(4i arctan(((√5)/6)))
wehave6+i5∣=36+5=416+i5=41eiarctan(56)(46+i5)4=44(41)4e4iarctan(56)=44(41)4e4iarctan(56)

Leave a Reply

Your email address will not be published. Required fields are marked *