Use these complex numbers from the table to write the following expression as a complex number in standard form. 2i(A*E) - C +G
( A = -3 - 4i, C = 13+i, E = 8 - 8i, and G = 2 + 12i)

=  5 − 101i

Step-by-step explanation:

2i(A ∙ E) − C + G

=

2i(-3 − 4i)(8 − 8i) − (13 + i) + (2 + 12i)

=

2i(-24 + 24i − 32i + 32i2) − (13 + i) + (2 + 12i)

=

2i(-24 − 8i − 32) − (13 + i) + (2 + 12i)

=

2i(-56 − 8i) − (13 + i) + (2 + 12i)

=

(-112i − 16i2) − (13 + i) + (2 + 12i)

=

(16 − 112i) − (13 + i) + (2 + 12i)

=

(16 − 112i) + (-13 − i) + (2 + 12i)

=

(16 − 13 + 2) + (-112i − i + 12i)

=

(16 − 13 + 2) + (-112 − 1 + 12)i

=  5 − 101i

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

is there a picture

step-by-step explanation: