How do you multiply #(-2-3i)(7+7i)#?

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Answer 1 · 2016-12-07T15:50:51

The answer is #=7-35i#

If 2 complex numbers are

#z_1=a_1+ib_1#

and #z_2=a_2+ib_2#

#i^2=-1#

then,

#z_1*z_2=(a_1+ib_1)(a_2+ib_2)#

#=a_1a_2+ia_1b_2+ia_2b_1+i^2b_1b_2#

#=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)#

So we have,

#(-2-3i)(7+7i)#

#=-14-14i-21i-21i^2#

#=-14+21-35i#

#=7-35i#