Answer to Question #138261 in Complex Analysis for kevin

Question

(a) Let z1 = 1 + √2i and 1 −√2i.
(i) Determine the polar form of z1.
(ii) Determine that the polar form of z2.
(iii) Use the polar forms of z1 and z2 to verify that z1 · z2 = 3
(iv) Use the polar forms of z1 and z2 to verify that −1/3 +2/3√2i =z1/z2

Answer 1

$\displaystyle(a) \\ z_1 = 1 + i\sqrt{2}, z_2 = 1 - i\sqrt{2}\\ (i) \\ \textsf{Polar form of}\, z_1 \\ \begin{aligned} r &= \sqrt{1 + (\sqrt{2})^2}\\ &=\sqrt{1 + 2} \\ &= \sqrt{3} \end{aligned} \\ \begin{aligned} \theta = \arctan\left(\frac{\sqrt{2}}{1}\right) &= \arctan(\sqrt{2})\\ &= 54.74\degree \end{aligned} \\ \therefore z_1 = \sqrt{3}(\cos(54.74\degree) + i\sin(54.74\degree)) \\ (ii)\\ \textsf{Polar form of}\, z_2\\ \begin{aligned} r &= \sqrt{1 + (\sqrt{2})^2}\\ &=\sqrt{1 + 2} \\ &= \sqrt{3} \end{aligned} \\ \textsf{Since}\, \theta\,\textsf{is in the fourth quadrant,}\\ \begin{aligned} \theta &=360\degree - \arctan\left(\frac{\sqrt{2}}{1}\right) \\&= 360\degree - \arctan(\sqrt{2}) \\&= 360\degree - 54.74\degree \\&= 305.26\degree \end{aligned}\\ \therefore z_2 = \sqrt{3}(\cos(305.26\degree) + i\sin(305.26\degree)) \\ (iii)\\\begin{aligned} z_1 \cdot z_2 &= \sqrt{3}(\cos(54.74\degree) + i\sin(54.74\degree)) \\&\times \sqrt{3}(\cos(305.26\degree) + i\sin(305.26\degree)) \\&= 3(\cos(54.74\degree)\cos(305.26\degree) \\&+ i(\cos(54.74\degree)\sin(305.26\degree) + \sin(54.74\degree)\cos(305.26\degree)) \\&- \sin(54.74\degree)\sin(305.26\degree)) \end{aligned}\\ \begin{aligned} \cos(305.26\degree) &= \cos(360\degree - 54.74\degree) = \cos(54.74\degree) \\ \sin(305.26\degree) &= -\sin(360\degree - 54.74\degree) = -\sin(54.74\degree) \end{aligned}\\ \begin{aligned} z_1 \cdot z_2 &= 3(\cos(54.74\degree)\cos(54.74\degree) \\&+ i(\cos(54.74\degree)\cdot -\sin(54.74\degree) + \sin(54.74\degree)\cos(54.74\degree)) \\&- \sin(54.74\degree)\cdot-\sin(54.74\degree)) \\&= 3(\cos^2(54.74\degree) + \sin^2(54.74\degree)) = 3\times 1 = 3 \end{aligned}\\ (iv) \\ \begin{aligned} \frac{z_1}{z_2} &= \frac{\sqrt{3}(\cos(54.74\degree) + i\sin(54.74\degree))}{\sqrt{3}(\cos(305.26\degree) + i\sin(305.26\degree))} \\&= \frac{(\cos(54.74\degree) + i\sin(54.74\degree))}{(\cos(305.26\degree) + i\sin(305.26\degree))} \times \frac{(\cos(305.26\degree) - i\sin(305.26\degree))}{(\cos(305.26\degree) - i\sin(305.26\degree))} \\&=\frac{\cos(54.74\degree)\cos(305.26\degree)}{{(\cos^2(305.26\degree) + \sin^2(305.26\degree)}}+\\& \frac{i(-\cos(54.74\degree)\sin(305.26\degree) + \sin(54.74\degree)\cos(305.26\degree))}{(\cos^2(305.26\degree) + \sin^2(305.26\degree)} \\&+ \frac{\sin(54.74\degree)\sin(305.26\degree)}{(\cos^2(305.26\degree) + \sin^2(305.26\degree)} \\&=\frac{\cos^2(54.74\degree) + i(\cos(54.74\degree)\sin(54.74\degree))}{1} \\&+ \frac{\sin(54.74\degree)\cos(54.74\degree)) - \sin^2(54.74\degree)}{1} \\&= \cos^2(54.74\degree) - \sin^2(54.74\degree) + 2i(\cos(54.74\degree)\sin(54.74\degree)) \\&= \cos^2(\arctan(\sqrt{2})) - \sin^2(\arctan(\sqrt{2})) \\&\hspace{4.5cm}+ 2i(\cos(\arctan(\sqrt{2}))\sin(\arctan(\sqrt{2}))) \end{aligned}\\ \textsf{Let}\, y = \cos(\arctan(\sqrt{x})), u = \arctan(\sqrt{x}) \\ y = \cos{u}, x = \tan{u} \\ 1 + \tan^2{u} = \sec^2{u} \implies \cos{u} = \frac{1}{\sqrt{1 + \tan^2{u}}}\\ \therefore y = \cos(\arctan(\sqrt{x})) = \frac{1}{\sqrt{1 + \tan^2{u}}}= \frac{1}{\sqrt{1 + x^2}}\\ \therefore \cos^2(\arctan(\sqrt{2})) = \frac{1}{1 + (\sqrt{2})^2} = \frac{1}{3} \, \& \\ \sin^2(\arctan(\sqrt{2})) = 1 - \cos^2(\arctan(\sqrt{2})) = 1 - \frac{1}{3} = \frac{2}{3} \\ \begin{aligned} \therefore \frac{z_1}{z_2} &= \frac{1}{3} - \frac{2}{3} + 2i\left(\sqrt{\frac{2}{3}\cdot\frac{1}{3}}\right) \\&= -\frac{1}{3} + i\frac{2}{3}\sqrt{2} \end{aligned}$