1. **Problem (a):** Given $u = 3 - 7i$ and $\overline{u}$ is the complex conjugate of $u$, find $2\overline{u} + 5iu$ in the form $a + bi$. 2. The complex conjugate $\overline{u}$ of $u = 3 - 7i$ is $3 + 7i$. 3. Calculate $2\overline{u}$: $$2\overline{u} = 2(3 + 7i) = 6 + 14i$$ 4. Calculate $5iu$: $$5iu = 5i(3 - 7i) = 5(3i - 7i^2) = 15i - 35i^2$$ 5. Since $i^2 = -1$, substitute: $$15i - 35(-1) = 15i + 35 = 35 + 15i$$ 6. Add the two results: $$2\overline{u} + 5iu = (6 + 14i) + (35 + 15i) = (6 + 35) + (14i + 15i) = 41 + 29i$$ --- 1. **Problem (b):** Use de Moivre's theorem to find $(3 - \sqrt{3}i)^8$ in the form $a + bi$. 2. Express the complex number in polar form: $$z = 3 - \sqrt{3}i$$ 3. Calculate modulus $r$: $$r = \sqrt{3^2 + (-\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}$$ 4. Calculate argument $\theta$: $$\theta = \arctan\left(\frac{-\sqrt{3}}{3}\right) = \arctan\left(-\frac{\sqrt{3}}{3}\right) = -30^\circ$$ 5. Using de Moivre's theorem: $$(r(\cos \theta + i \sin \theta))^8 = r^8 (\cos 8\theta + i \sin 8\theta)$$ 6. Calculate $r^8$: $$r^8 = (2\sqrt{3})^8 = (2^8)(\sqrt{3})^8 = 256 \times 3^4 = 256 \times 81 = 20736$$ 7. Calculate $8\theta$: $$8 \times (-30^\circ) = -240^\circ$$ 8. Find $\cos(-240^\circ)$ and $\sin(-240^\circ)$: $$\cos(-240^\circ) = \cos(240^\circ) = -\frac{1}{2}$$ $$\sin(-240^\circ) = -\sin(240^\circ) = -\left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}$$ 9. Substitute back: $$z^8 = 20736 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 20736 \times -\frac{1}{2} + 20736 \times i \frac{\sqrt{3}}{2} = -10368 + 10368 \sqrt{3} i$$ --- 1. **Problem (c):** Given $w = a(\cos 160^\circ + i \sin 160^\circ)$, $v = b(\cos \theta + i \sin \theta)$, $|w| = 5$, and $wv = -15$, find one possible set of $a$, $b$, and $\theta$. 2. Since $|w| = a = 5$. 3. The product $wv$ in polar form is: $$wv = ab \left(\cos(160^\circ + \theta) + i \sin(160^\circ + \theta)\right)$$ 4. Given $wv = -15$, which is a real number with imaginary part zero, so: $$ab \sin(160^\circ + \theta) = 0$$ 5. Also, the real part: $$ab \cos(160^\circ + \theta) = -15$$ 6. Since $a = 5$, substitute: $$5b \sin(160^\circ + \theta) = 0 \implies \sin(160^\circ + \theta) = 0$$ 7. The sine is zero at angles $k \times 180^\circ$, so: $$160^\circ + \theta = k \times 180^\circ$$ 8. For $k=1$: $$160^\circ + \theta = 180^\circ \implies \theta = 20^\circ$$ 9. Substitute $\theta = 20^\circ$ into the cosine equation: $$5b \cos(180^\circ) = -15$$ 10. Since $\cos(180^\circ) = -1$: $$5b (-1) = -15 \implies -5b = -15 \implies b = 3$$ **Final answers:** - (a) $41 + 29i$ - (b) $-10368 + 10368 \sqrt{3} i$ - (c) $a = 5$, $b = 3$, $\theta = 20^\circ$