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Answer:
(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.
