Question

w=(3-i)/(2i-1) solving this i get that w= -i-1 now i want to express w in polar form: r=sqrt2

but can you explain why arg(w)= -3pi/4 ? when arctan(1) = pi/4 (how do they get -3pi/4)

and if arg(w)= -3pi/4 can i than write 5pi/4 instead

and z^4=w and how to draw the solutions in the complex plane

Answer 1

Answer on Question #44245, Math, Complex Analysis

$w = (3 - i) / (2i - 1)$ solving this i get that $w = -i - 1$ now i want to express w in polar form: $r = \mathrm{sqrt}2$ but can you explain why $\arg (w) = -3pi / 4$ ? when $\arctan (1) = pi / 4$ (how do they get -3pi/4) and if $\arg (w) = -3pi / 4$ can i than write $5pi / 4$ instead and $z^{\wedge}4 = w$ and how to draw the solutions in the complex plane.

Solution.

$$w = \frac {3 - i}{2 i - 1} = \frac {(3 - i) (- 2 i - 1)}{(2 i - 1) (- 2 i - 1)} = \frac {- 6 i - 3 - 2 + i}{1 + 4} = \frac {- 5 - 5 i}{5} = - 1 - i.$$

The modulus equals $|w| = \sqrt{1^2 + 1^2} = \sqrt{2}$ . Hence $w = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i\right)$ . The argument of $z$ is the value of $\phi$ for which $\cos \phi = -\frac{1}{\sqrt{2}}$ and $\sin \phi = -\frac{1}{\sqrt{2}}$ . The $\frac{\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}} = \tan \left(-\frac{3\pi}{4}\right) = \frac{\sin \left(-\frac{3\pi}{4}\right)}{\cos \left(-\frac{3\pi}{4}\right)} = 1$ , but $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \neq -\frac{1}{\sqrt{2}}$ . The set of values $\phi$ is $\left\{-\frac{3\pi}{4} + 2\pi n: n \in \mathbb{Z}\right\}$ . To make $\arg w$ a well-defined function, it is defined as a function which equals the value of $\phi$ in the open-closed (closed-open) interval of the length $2\pi$ . Authors often use $(-\pi; \pi]$ , but in some literature you may find interval $(0; 2\pi]$ , the value in this interval is often denoted as $\operatorname{Arg}(w)$ . Hence you may write that $w = \sqrt{2}\left(\cos \frac{5\pi}{4} + \sin \frac{5\pi}{4}\right)$ , but $\arg (w) \neq \frac{5\pi}{4}$ .

From the $n$ -th root formula the solution of the equation $z^4 = w$ .

$$z _ {0} = \sqrt [ n ]{2} \left(\sin \left(- \frac {3 \pi}{1 6}\right) + i \cos \left(- \frac {3 \pi}{1 6}\right)\right);$$ $$z _ {1} = \sqrt [ n ]{2} \left(\sin \left(\frac {- 3 \pi}{4} + 2 \pi\right) + i \cos \left(\frac {- 3 \pi}{4} + 2 \pi\right)\right) = \sqrt [ n ]{2} \left(\sin \left(\frac {5 \pi}{1 6}\right) + i \cos \left(\frac {5 \pi}{1 6}\right)\right);$$ $$z _ {2} = \sqrt [ n ]{2} \left(\sin \left(\frac {- 3 \pi}{4} + 4 \pi\right) + i \cos \left(\frac {- 3 \pi}{4} + 4 \pi\right)\right) = \sqrt [ n ]{2} \left(\sin \left(\frac {1 3 \pi}{1 6}\right) + i \cos \left(\frac {1 3 \pi}{1 6}\right)\right);$$ $$\begin{array}{l} z _ {3} = \sqrt [ n ]{2} \left(\sin \left(\frac {- 3 \pi}{4} + 6 \pi\right) + i \cos \left(\frac {- 3 \pi}{4} + 6 \pi\right)\right) = \sqrt [ n ]{2} \left(\sin \left(\frac {2 1 \pi}{1 6}\right) + i \cos \left(\frac {2 1 \pi}{1 6}\right)\right) \\ = \sqrt [ n ]{2} \left(\sin \left(- \frac {1 1 \pi}{1 6}\right) + i \cos \left(- \frac {1 1 \pi}{1 6}\right)\right). \\ \end{array}$$

The roots of the equation $z^4 = w$ lie in the vertices of the squares with a diagonal $2|z| = 2^{\frac{n}{2}}2$ and center at $(0;0)$ . The square is rotated at angle $\frac{5\pi}{16}$ on the center of the square.

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