Here are the solutions for Question 3. 3A. Given $Z_1 = 2+4i$ and $Z_2 = 1+i$. i) Product $Z_1 Z_2$ Step 1: Multiply the complex numbers. $$ Z_1 Z_2 = (2+4i)(1+i) $$ Step 2: Expand the product. $$ (2+4i)(1+i) = 2(1) + 2(i) + 4i(1) + 4i(i) $$ $$ = 2 + 2i + 4i + 4i^2 $$ Step 3: Simplify using $i^2 = -1$. $$ 2 + 6i + 4(-1) = 2 + 6i - 4 = -2 + 6i $$ The product is $\boxed{-2 + 6i}$. ii) Division $Z_1 / Z_2$ Step 1: Write the division as a fraction. $$ \frac{Z_1}{Z_2} = \frac{2+4i}{1+i} $$ Step 2: Multiply the numerator and denominator by the conjugate of the denominator, which is $1-i$. $$ \frac{2+4i}{1+i} = \frac{(2+4i)(1-i)}{(1+i)(1-i)} $$ Step 3: Expand the numerator and denominator. Numerator: $(2+4i)(1-i) = 2(1) + 2(-i) + 4i(1) + 4i(-i) = 2 - 2i + 4i - 4i^2 = 2 + 2i - 4(-1) = 2 + 2i + 4 = 6 + 2i$ Denominator: $(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$ Step 4: Combine the results and simplify. $$ \frac{6+2i}{2} = \frac{6}{2} + \frac{2i}{2} = 3 + i $$ The division is $\boxed{3 + i}$. 3B. Given $Z_1 = 1-2i$ and $Z_2 = 1+2i$. i) Product $Z_1 Z_2$ Step 1: Multiply the complex numbers. $$ Z_1 Z_2 = (1-2i)(1+2i) $$ Step 2: Use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$. $$ 1^2 - (2i)^2 $$ Step 3: Simplify using $i^2 = -1$. $$ 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5 $$ The product is $\boxed{5}$. ii) Division $Z_1 / Z_2$ Step 1: Write the division as a fraction. $$ \frac{Z_1}{Z_2} = \frac{1-2i}{1+2i} $$ Step 2: Multiply the numerator and denominator by the conjugate of the denominator, which is $1-2i$. $$ \frac{1-2i}{1+2i} = \frac{(1-2i)(1-2i)}{(1+2i)(1-2i)} $$ Step 3: Expand the numerator and denominator. Numerator: $(1-2i)(1-2i) = 1(1) + 1(-2i) - 2i(1) - 2i(-2i) = 1 - 2i - 2i + 4i^2 = 1 - 4i + 4(-1) = 1 - 4i - 4 = -3 - 4i$ Denominator: $(1+2i)(1-2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5$ Step 4: Combine the results and simplify. $$ \frac{-3-4i}{5} = -\frac{3}{5} - \frac{4}{5}i $$ The division is $\boxed{-\frac{3}{5} - \frac{4}{5}i}$.