MATHEMATICS 2 (MA10193) EXAMPLES SHEET 3: SOLUTIONS

1. (7 − 2i) + (8 + 5i) = 15 + 3i; (3 − i)(6 + 5i) = 23 + 9i; (3 − i)/(6 + 5i) = \frac{13−21i}{65}; (2 + i)/i = 1 − 2i; \sqrt{−i} = ±\frac{1+i}{\sqrt{2}}.

2.

A^{−1} = \frac{1}{3} ( \begin{matrix}1 & −1 & 1 \\ 5− & −2 & −1 \\ −7 & 4 & 2 \end{matrix} ) .

3. z^{2} = 8 + 6i. You should be able to draw the points (3,1) and (8,6) without help.

4. z_{1}z_{2} = (1 − 2i)(4 + yi) = 4 + yi − 8i + 2y, so the real part is 4 + 2y which is zero if and only if y = −2. z_{1}/z_{2} = (1 − 2i)/(4 + yi) = (1 − 2i)(4 − yi)/(16 + y^{2}). Since 16 + y^{2} is real, this is real if and only if the numerator, (1 − 2i)(4 − yi), is real; that is, if its imaginary part is zero. But the imaginary part of (1 − 2i)(4 − yi) = 4 + 2y − yi − 8i is −y − 8, so we need y = 8.

5. |z| = \sqrt{\sqrt{3}^{2} + 1^{2}} = 2, and tan arg z = 1/\sqrt{3} so arg z = π/3 (since z is in the first quadrant). Note: it is π/3 exactly and a calculator approximation gives the wrong answer. The smallest positive integer n such that z^{n} is a real number is 3, since arg z^{3} = 3 arg z = π (this is why an approximation is wrong: we need to get exactly π here, not something close to it).

6. If x^{2} − 5x + 32 = 0 then x = \frac{1}{2}(5 ± i\sqrt{103}). The eigenvalues of ( \begin{matrix}3 & 13 \\ −2 & 2 \end{matrix} ) are found

by solving the equation | \begin{matrix}3 − x & 13 \\ −2 & 2 − x \end{matrix} | = 0 and that is x^{2} − 5x + 32 = 0 so those are the eigenvalues.

GKS, 9/05/05