How do you find the real and imaginary part #12i^12 + pi(i)#?
1 Answer
Mar 22, 2016
Explanation:
Note that we can simplify
#{(i=color(red)(sqrt(-1))),(i^2=(color(red)(sqrt(-1)))^2=color(blue)(-1)),(i^4=(i^2)^2=(color(blue)(-1))^2=color(green)1),(i^12=(i^4)^3=color(green)1^3=1):}#
Thus, the expression simplifies to be:
#12i^12+pii=12(1)+pii=12+pii#
This is a complex number in the form
#a+bi#
where