How do you write #-8-(3+2i)-(9-4i)# in standard form?

1 Answer
Jan 6, 2018

#-20 + 2i#

Explanation:

Standard form for complex numbers is #a+bi#, where #a# and #b# are integers. To get this, we just have to combine all the terms with #i# and all the terms without #i#.

Simplifying the expression:

#-8-(3+2i)-(9-4i)#

#=-8-3-2i-9+4i#

Regrouping the terms:

#= (-8-3-9) + (-2i+4i)#

# = -20 + 2i#