Write the complex number #(2-3i)(-1+4i)# in standard form?

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Answer 1 · 2015-05-11T14:02:07

Multiply the two brackets as:
#(2-3i)(-1+4i)=-2+8i+3i-12i^2=#
But #i^2=(sqrt(-1))^2=-1#
so:
#=-2+11i+12=10+11i#