XC - PROJECT
Include i in multiplication
(a + bi)(c + di) == ac + adi + bci + bdi2
Of course, i2 == -1 , by definition
(a + bi)(c + di) == ac + adi + bci + bd(-1)
(a + bi)(c + di) == ac + adi + bci - bd
(a + bi)(c + di) == (ac – bd) + (ad+ bc)i
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