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

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

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Answer 1 · 2018-01-06T19:40:50

#-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#

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

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