If Z1 = 12 + 6݅ and Z2 = a + bi (where a, b ∈ R) are two complex number, we can say that the product Z1Z2is imaginary A) If and only if ܽ = ܾ = 0 B) If b = 2a C) If a = −2b D) for each value of a and b

A) if and only if a = 0

Step-by-step explanation:

Since Z1 = 12 + 6݅  and Z2 = a + bi , then the product, Z1Z2 = (12 + 6)(a + bi)

Z1Z2 = (12 + 6)(a + bi)

Expanding the brackets, we have

Z1Z2 = 12a + 12bi + 6a + 6bi

Collecting like terms, we have

Z1Z2 = 12a + 6a + 12bi +6bi

Z1Z2 = 12a + 6a + (12b +6b)i

Simplifying, we have

Z1Z2 = 18a + 18bi

For Z1Z2 to be imaginary, then the real part must be zero.

That is 18a = 0 ⇒ a = 0

So, Z1Z2 is imaginary if and only if a = 0

b

2

−

b

b

3

+

7

b

2

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14

b

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6

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−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation: