1. **Problem:** Given $z = -3 + 4i$ and $w = 5 + 2i$, express $w$ in the form $\alpha + bi$ where $\alpha$ and $b$ are real numbers. **Step 1:** Identify the real and imaginary parts of $w$. $w = 5 + 2i$ Here, the real part $\alpha = 5$ and the imaginary part $b = 2$. **Answer:** $w = 5 + 2i$. 2. **Problem:** Given $z = (\alpha + i)^4$ where $\alpha$ is real, find values of $\alpha$ such that: (a) $z$ is real. (b) $z$ is purely imaginary. **Step 1:** Expand $z = (\alpha + i)^4$ using the binomial theorem: $$(\alpha + i)^4 = \sum_{k=0}^4 \binom{4}{k} \alpha^{4-k} i^k$$ Calculate each term: - $k=0$: $\binom{4}{0} \alpha^4 i^0 = \alpha^4$ - $k=1$: $\binom{4}{1} \alpha^3 i = 4 \alpha^3 i$ - $k=2$: $\binom{4}{2} \alpha^2 i^2 = 6 \alpha^2 (i^2) = 6 \alpha^2 (-1) = -6 \alpha^2$ - $k=3$: $\binom{4}{3} \alpha i^3 = 4 \alpha (i^3) = 4 \alpha (-i) = -4 \alpha i$ - $k=4$: $\binom{4}{4} i^4 = 1 \cdot 1 = 1$ Sum real parts: $$\alpha^4 - 6 \alpha^2 + 1$$ Sum imaginary parts: $$(4 \alpha^3 - 4 \alpha) i = 4 \alpha (\alpha^2 - 1) i$$ So, $$z = (\alpha^4 - 6 \alpha^2 + 1) + 4 \alpha (\alpha^2 - 1) i$$ **Step 2:** For $z$ to be real, imaginary part must be zero: $$4 \alpha (\alpha^2 - 1) = 0$$ Solve: $$\alpha = 0 \quad \text{or} \quad \alpha^2 = 1 \Rightarrow \alpha = \pm 1$$ **Step 3:** For $z$ to be purely imaginary, real part must be zero: $$\alpha^4 - 6 \alpha^2 + 1 = 0$$ Let $x = \alpha^2$, then: $$x^2 - 6x + 1 = 0$$ Solve quadratic: $$x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2 \sqrt{2}$$ Since $x = \alpha^2$, $\alpha = \pm \sqrt{3 + 2 \sqrt{2}}$ or $\pm \sqrt{3 - 2 \sqrt{2}}$. 3. **Problem:** Given $\alpha + bi$ is the conjugate of $(\alpha + bi)^2$, find all possible pairs $(\alpha, b)$ where $\alpha, b$ are real. **Step 1:** Compute $(\alpha + bi)^2$: $$(\alpha + bi)^2 = \alpha^2 + 2 \alpha b i + (bi)^2 = \alpha^2 - b^2 + 2 \alpha b i$$ **Step 2:** The conjugate of $(\alpha + bi)^2$ is: $$\overline{(\alpha + bi)^2} = \alpha^2 - b^2 - 2 \alpha b i$$ **Step 3:** Given $\alpha + bi = \overline{(\alpha + bi)^2}$, equate real and imaginary parts: Real parts: $$\alpha = \alpha^2 - b^2$$ Imaginary parts: $$b = -2 \alpha b$$ **Step 4:** From imaginary parts: $$b = -2 \alpha b \Rightarrow b + 2 \alpha b = b(1 + 2 \alpha) = 0$$ So either: - $b = 0$, or - $1 + 2 \alpha = 0 \Rightarrow \alpha = -\frac{1}{2}$ **Step 5:** Case 1: $b=0$ From real parts: $$\alpha = \alpha^2 - 0 = \alpha^2$$ So: $$\alpha^2 - \alpha = 0 \Rightarrow \alpha(\alpha - 1) = 0$$ Thus, $\alpha = 0$ or $\alpha = 1$. Pairs: $(0,0)$ and $(1,0)$ **Step 6:** Case 2: $\alpha = -\frac{1}{2}$ From real parts: $$-\frac{1}{2} = \left(-\frac{1}{2}\right)^2 - b^2 = \frac{1}{4} - b^2$$ Solve for $b^2$: $$-\frac{1}{2} - \frac{1}{4} = -b^2 \Rightarrow -\frac{3}{4} = -b^2 \Rightarrow b^2 = \frac{3}{4}$$ So: $$b = \pm \frac{\sqrt{3}}{2}$$ Pairs: $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ 4. **Problem:** Simplify and write in the form $\alpha + bi$: (a) $\frac{1}{3 + 2i} + \frac{1}{3 - 2i}$ (b) $\frac{3 + i}{3 - i} - 4$ (c) $\frac{3 - 2i}{1 - i^2}$ **(a) Step 1:** Find common denominator and simplify each term. Recall $\frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2}$. Calculate: $$\frac{1}{3 + 2i} = \frac{3 - 2i}{3^2 + 2^2} = \frac{3 - 2i}{9 + 4} = \frac{3 - 2i}{13}$$ $$\frac{1}{3 - 2i} = \frac{3 + 2i}{13}$$ Sum: $$\frac{3 - 2i}{13} + \frac{3 + 2i}{13} = \frac{(3 - 2i) + (3 + 2i)}{13} = \frac{6}{13}$$ Answer: $\frac{6}{13} + 0i$ **(b) Step 1:** Simplify $\frac{3 + i}{3 - i}$. Multiply numerator and denominator by conjugate of denominator: $$\frac{3 + i}{3 - i} \times \frac{3 + i}{3 + i} = \frac{(3 + i)^2}{3^2 - (-i)^2} = \frac{(3 + i)^2}{9 + 1} = \frac{(3 + i)^2}{10}$$ Calculate numerator: $$(3 + i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i$$ So: $$\frac{8 + 6i}{10} = \frac{8}{10} + \frac{6}{10} i = \frac{4}{5} + \frac{3}{5} i$$ Step 2: Subtract 4: $$\left(\frac{4}{5} + \frac{3}{5} i\right) - 4 = \frac{4}{5} - 4 + \frac{3}{5} i = -\frac{16}{5} + \frac{3}{5} i$$ **(c) Step 1:** Simplify denominator: $$1 - i^2 = 1 - (-1) = 1 + 1 = 2$$ Step 2: Simplify numerator: $$3 - 2i$$ Step 3: Divide numerator by denominator: $$\frac{3 - 2i}{2} = \frac{3}{2} - i$$ **Final answers:** (a) $\frac{6}{13}$ (b) $-\frac{16}{5} + \frac{3}{5} i$ (c) $\frac{3}{2} - i$