How do you write the expression #(sqrt(2) - i)^6# in the standard form a + bi?
1 Answer
Mar 5, 2016
Explanation:
Calling
We can express
With
And, since
=>
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)#