How do you write the complex number #(4-2i)(-3+i)# in standard form?
1 Answer
May 7, 2018
# (4-2i)(-3+i) = -10+10i#
Explanation:
We seek a simplification of:
# (4-2i)(-3+i) #
Now::
# (4-2i)(-3+i) = 4(-3+i) -2i(-3+i) #
# " " = -12+4i+6i-2i^2 #
# " " = -12+10i-2i^2 #
# " " = -12+10i-2(-1) \ \ \ \ because i^2=-1#
# " " = -12+10i+2#
# " " = -10+10i#
