Sketch a proof that every complex number has an additive inverse?
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additive inverse of a+bi is -a-bi
depends on which form of complex number you use. The definition of a complex number is the ordered pair (a,b), where a,b are real numbers. so the additive inverse of (a,b) is (-a,-b) since (a,b) + (-a,-b) (a-a,b-b) (0,0) If you use the equivalent form a + bi, then its additive inverse is -a - bi. You could go ahead and apply addition rule for real numbers and do a + bi - a - bi (a-a) + (bi - bi) 0 + 0i, but I prefer to use ordered pairs. We define i (0,1) and (a,b) (c,d) (ad - bd, ac + bd). first note that bi (b,0) (0,1) (b,0) we have: (a + bi) + (-a - bi) (a,0) + (b,0) + (-a,0) + (-b,0) (a + b - a - b, 0 + 0 + 0 + 0) (0,0)
