find the additive inverse 3-4i÷2-3i
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3-4i/2-3i*2+3i/2+3i
(3-4i)(2+3i)/4+9
(6+9i-8i+12)/13
(18+i)/13
=18/13+1/13*i
additive inverse=-18/13-i/13
(3-4i)(2+3i)/4+9
(6+9i-8i+12)/13
(18+i)/13
=18/13+1/13*i
additive inverse=-18/13-i/13
vsk96garg:
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