Problem Show that $$\frac{-1 + \sqrt{3}i}{2}$$ is a cube root of 1 (meaning that it’s cube equals 1). Solution One solution is to take the cube of the given complex number and observe that it equals one. $$\left(\frac{-1 + \sqrt{3}i}{2}\right)^3 = \frac{1}{8}\left(-1 + \sqrt{3}i\right)^3 $$ $$= \frac{1}{8}\left(-1 + 3\sqrt{3}i - 3(-3) -3\sqrt{3}i\right) = \frac{1}{8} \cdot 8 = 1$$ Another is to notice that $\cos(2\pi/3) = -1/2$ and that $\sin(2\pi/3) = i \sqrt{3} / 2$, which by eulers formula implies: $$\sqrt[3]{1} = \sqrt[3]{e^{2\pi i}} = e^{\frac{2\pi}{3}i} = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = \frac{-1 + \sqrt{3}i}{2}$$

Problem Suppose aand bare real numbers, not both 0. Find real numbers cand dsuch that $$\frac{1}{a + bi} = c + di$$ Solution A standard trick when dealing with the reciprocal of a complex numbers is to multiply the numerator and the denominator by it’s complex conjugate, as this will make the denominator real. Applying this trick: $$c + di = \frac{1}{a + bi} = \frac{a - bi}{a - bi} \cdot \frac{1}{a + bi} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i$$ From comparing the real and imaginary parts of the left and right hand sides above: $$c = \frac{a}{a^2 + b^2} \land d= \frac{b}{a^2 + b^2}$$ Note that the solution lies on the same line as $\overline{z}$ and it’s magnitude is the reciprocal of the magnitude of $z = a + bi$....