How do you write the expression #(sqrt(2) - i)^6# in the standard form a + bi?

1 Answer
Mar 5, 2016

#-23+i10sqrt(2)~=-23+i*14.142136#

Explanation:

Calling
#C=sqrt(2)-i#
We can express #C# as #|C| /_ phi#
With
#|C|=sqrt((sqrt(2))^2+1^2)=sqrt(2+1)=sqrt(3)#
And, since #-90^@< phi<0#
#phi=-tan^(-1)(1/sqrt(2))=-35.26439^@#
=> #C=sqrt(3)" " /_-35.26439^@#

Then

#C^6=(sqrt(3))^6" "/_6*(-35.26439^@)#
#C^6=27" "/_-211.58634^@#
#C^6=27cos(-211.58634^@)+i*27sin(-211.58634^@)#
#C^6=-23+i14.142136#
If you like, reminding that #sqrt(2)=1.4142136#
#C^6=-23+i10sqrt(2)#