Suppose a + bi and c + di are complex numbers with b not equaling 0. (a) Calculate (a + bi) + (c + di) and (a + bi)(c + di).
(b) Show that if both the sum and product are real numbers, then either the complex numbers are real numbers or conjugates.

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  (a) sum: (a+c) +(b+d)i

     product: (ac -bd) +(bc +ad)i

  (b) (b+d)=0 and (bc+ad)=0 ⇒ a=c, d=-b or b=d=0

Step-by-step explanation:

(a) Combining like terms the sum is ...

  (a +bi) +(c +di) = (a+c) +(b+d)i . . . . sum

And the product is ...

  (a+bi)(c+di) = ac +(ad+bc)i +bd·i²

Since i = √-1, i² = -1 and the product can be written as ...

  (a+bi)(c+di) = (ac-bd) +(ad+bc)i . . . . product

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(b) If both the sum and product are real numbers, then we have ...

  b +d = 0

  ad +bc = 0

The first equation tells us d = -b. Substituting that into the second equation, we get ...

  a(-b) +b(c) = 0

  b(c -a) = 0

The zero product rule tells us this will be true if and only if b = 0 or c = a.

  if b = 0, then d = 0 and both numbers are real.

  if c = a, then c+di = a-bi and the numbers are conjugates.

Hence, if both the sum and product are real, both are real numbers or they are conjugates.

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