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9709 P32 - Nov 2021 - Q3
1935
(a) Given the complex numbers \(u = a + ib\) and \(w = c + id\), where \(a, b, c\) and \(d\) are real, prove that \((u + w)^* = u^* + w^*\).
(b) Solve the equation \((z + 2 + i)^* + (2 + i)z = 0\), giving your answer in the form \(x + iy\) where \(x\) and \(y\) are real.
Solution
(a) Substitute \(u = a + ib\) and \(w = c + id\). The conjugate of \(u + w\) is \((a + c) + i(b + d)\)^* = (a + c) - i(b + d)\). The conjugate of \(u\) is \(a - ib\) and the conjugate of \(w\) is \(c - id\). Therefore, \(u^* + w^* = (a - ib) + (c - id) = (a + c) - i(b + d)\). Thus, \((u + w)^* = u^* + w^*\).
(b) Let \(z = x + iy\). Then \((z + 2 + i)^* = (x + 2 + i(y + 1))^* = x + 2 - i(y + 1)\). Substitute into the equation: \(x + 2 - i(y + 1) + (2 + i)(x + iy) = 0\). Simplify: \(x + 2 - i(y + 1) + (2x - y) + i(2y + x) = 0\). Equate real and imaginary parts to zero: \(3x - y + 2 = 0\) and \(x + y - 1 = 0\). Solve these equations: From \(x + y - 1 = 0\), we get \(y = 1 - x\). Substitute into \(3x - y + 2 = 0\): \(3x - (1 - x) + 2 = 0\) gives \(4x + 1 = 0\), so \(x = -\frac{1}{4}\). Then \(y = 1 - (-\frac{1}{4}) = \frac{5}{4}\). Therefore, \(z = -\frac{1}{4} + \frac{5}{4}i\).
