1. Problem 1: Perform operations on complex numbers.\n(i) Add (3,2) + (9,3):\n\nAdd the real parts: $3 + 9 = 12$.\nAdd the imaginary parts: $2 + 3 = 5$.\nAnswer: $12 + 5i$.\n\n(ii) Add (3 2) + (2 3):\n\nThis seems to be notation for complex numbers $(3 + 2i) + (2 + 3i)$.\nAdd real parts: $3 + 2 = 5$.\nAdd imaginary parts: $2 + 3 = 5$.\nAnswer: $5 + 5i$.\n\n(iii) Subtract (15,12) - (10,-9):\n\nSubtract real parts: $15 - 10 = 5$.\nSubtract imaginary parts: $12 - (-9) = 12 + 9 = 21$.\nAnswer: $5 + 21i$.\n\n(iv) Subtract (4 8) - (4 6):\n\nInterpreted as $(4 + 8i) - (4 + 6i)$.\nReal parts: $4 - 4 = 0$.\nImaginary parts: $8 - 6 = 2$.\nAnswer: $0 + 2i$.\n\n(v) Multiply (1,2)(1,-2):\n\nLet the numbers be $1 + 2i$ and $1 - 2i$.\nMultiply using distributive property:\n$ (1)(1) + (1)(-2i) + (2i)(1) + (2i)(-2i) = 1 - 2i + 2i -4i^2 = 1 + 0 + 4$ (since $i^2 = -1$).\nAnswer: $5 + 0i = 5$.\n\n(vi) Multiply (4,-5)(5,-4):\n\nNumbers $4 - 5i$ and $5 - 4i$.\nMultiply:\n$4*5 + 4*(-4i) + (-5i)*5 + (-5i)*(-4i) = 20 - 16i -25i + 20i^2 = 20 - 41i - 20$ (since $i^2=-1$).\nSimplify real parts: $20 - 20 = 0$.\nAnswer: $0 - 41i$.\n\n(vii) Divide (3,-7) ÷ (3,2):\n\nNumbers: $3 -7i$ divided by $3 + 2i$.\nMultiply numerator and denominator by conjugate of denominator $3 - 2i$: denominator becomes\n$ (3 + 2i)(3 - 2i) = 9 - 6i + 6i -4i^2 = 9 + 4 = 13 $.\nNumerator:\n$ (3 - 7i)(3 - 2i) = 9 - 6i -21i + 14i^2 = 9 - 27i -14 = (9 -14) - 27i = -5 - 27i$.\nDivide each part by 13:\n$\frac{-5}{13} - \frac{27}{13}i$.\nAnswer: $-\frac{5}{13} - \frac{27}{13}i$.\n\n(viii) Divide (4,5) ÷ (2,-3):\n\nNumbers $4 + 5i$ and $2 - 3i$.\nMultiply numerator and denominator by conjugate of denominator $2 + 3i$.\nDenominator: $ (2 - 3i)(2 + 3i) = 4 + 6i - 6i -9i^2 = 4 + 9 = 13$.\nNumerator:\n$ (4 + 5i)(2 + 3i) = 8 + 12i + 10i + 15i^2 = 8 + 22i - 15 = (8 - 15) + 22i = -7 + 22i$.\nDivide by 13:\n$-\frac{7}{13} + \frac{22}{13}i$.\nAnswer: $-\frac{7}{13} + \frac{22}{13}i$.\n\n2. Problem 2: Simplify and write in form $a+ib$.\n(i) $\frac{-1}{1+i}$:\nMultiply numerator and denominator by conjugate $1 - i$:\n$\frac{-1}{1+i} \cdot \frac{1 - i}{1 - i} = \frac{-1(1 - i)}{1 - i^2} = \frac{-1 + i}{1 - (-1)} = \frac{-1 + i}{2} = -\frac{1}{2} + \frac{1}{2}i$.\nAnswer: $-\frac{1}{2} + \frac{1}{2}i$.\n\n(ii) $(1+i)^4$:\nFirst find $(1+i)^2 = (1)^2 + 2*i*1 + i^2 = 1 + 2i -1 = 2i$.\nThen $(1+i)^4 = ((1+i)^2)^2 = (2i)^2 = 4i^2 = 4(-1) = -4$.\nAnswer: $-4 + 0i$.\n\n(iii) $\left(\frac{1}{1+i}\right)^2$:\nFrom (i), $\frac{1}{1+i} = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}i$.\nSquare:\n$\left(\frac{1}{2} - \frac{1}{2}i\right)^2 = \frac{1}{4} - 2*\frac{1}{2}*\frac{1}{2}i + \frac{1}{4}i^2 = \frac{1}{4} - \frac{1}{2}i + \frac{1}{4}(-1)= \frac{1}{4} - \frac{1}{4} - \frac{1}{2}i = -\frac{1}{2}i$.\nAnswer: $0 - \frac{1}{2}i$.\n\n(iv) $(1+i)^8$:\nUsing $(1+i)^4 = -4$ from (ii), then $(1+i)^8 = ((1+i)^4)^2 = (-4)^2 = 16$.\nAnswer: $16 + 0i$.\n\n3. Problem 3: Verify given equalities for $z_1 = 4 + 6i$ and $z_2 = \frac{1}{2} - 2i$.\n(i) Verify $z_1 + z_2 = z_1 + z_2$:\nCompute sum:\n$4 + 6i + \frac{1}{2} - 2i = 4 + \frac{1}{2} + (6i - 2i) = \frac{9}{2} + 4i$.\nBoth sides match by simple addition, so equality holds.\n\n(ii) Verify $z_1 - z_2 = z_1 - z_2$:\nCompute difference:\n$4 + 6i - \left(\frac{1}{2} - 2i\right) = 4 + 6i - \frac{1}{2} + 2i = \frac{7}{2} + 8i$.\nBoth sides match, equality holds.\n\n4. Problem 4: Given $z_1 = 1 + i$ and $z_2 = 1 - i$, verify\n(i) $\frac{z_1}{z_2} = \frac{z_1}{z_2}$ (trivially true). Compute:\n\nMultiply numerator and denominator by conjugate of denominator $1 + i$:\n$\frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i -1}{2} = \frac{2i}{2} = i$.\nAnswer: $0 + 1i$.\nThis shows the expression on either side equals $i$.\n\n(ii) Verify $\frac{z_1}{z_2} = \frac{z_1}{z_2}$ (same as above), which is true and equals $i$.\n