Question

Obtain the polar and exponential representations of z_1 , z_2 and z_1 z_2 , where z_1. = 1/2 - ( 2i) and z_2 = 3+i ?

Answer 1

Answer on Question #74588 - Math / Complex Analysis - for completion.

Obtain the polar and exponential representations of $z_{1}, z_{2}$ and $z_{1}z_{2}$, where $z_{1} = \frac{1}{2} - 2i$ and $z_{2} = 3 + i$.

**Solution:**

$$z_{1} = \frac{1}{2} - 2i;$$ $$|z_{1}| = \sqrt{x^{2} + y^{2}}, \text{ where } x = Re\, z_{1} = \frac{1}{2}, y = Im\, z_{1} = -2;$$ $$arg\, z_{1} = \varphi; \varphi = arc \tan \frac{y}{x},$$ $$|z_{1}| = \sqrt{\left(\frac{1}{2}\right)^{2} + (-2)^{2}} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{1 + 16}{4}} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \approx 2.0616;$$ $$\varphi = arc \tan \frac{-2}{\frac{1}{2}} = arc \tan(-4);$$

**Polar and exponential representation of $z_{1}$:**

$$z_{1} = \frac{\sqrt{17}}{2} (\cos(arc \tan(-4)) + \sin(arc \tan(-4)));$$ $$z_{1} = \frac{\sqrt{17}}{2} e^{iarc \tan(-4)};$$ $$z_{2} = 3 + i;$$ $$|z_{2}| = \sqrt{x^{2} + y^{2}}, \text{ where } x = Re\, z_{2} = 3, y = Im\, z_{2} = 1;$$ $$arg\, z_{2} = \varphi; \varphi = arc \tan \frac{y}{x},$$ $$|z_{2}| = \sqrt{(3)^{2} + (1)^{2}} = \sqrt{9 + 1} = \sqrt{10} \approx 3.1623;$$ $$\varphi = arc \tan \frac{1}{3};$$

**Polar and exponential representation of $z_{2}$:**

$$z_{2} = \sqrt{10} \left(\cos\left(arc \tan \frac{1}{3}\right) + \sin(arc \tan \frac{1}{3})\right);$$ $$z_{2} = \sqrt{10} e^{iarc \tan \frac{1}{3}};$$ $$z_{1}z_{2} = (x_{1} + iy_{1})(x_{2} + iy_{2}) = (x_{1}x_{2} - y_{1}y_{2}) + i(y_{1}x_{2} + x_{1}y_{2});$$ $$\begin{array}{l} z_{1} z_{2} = \left(\frac{1}{2} - 2i\right) (3 + i) = \left(\frac{1}{2} * 3 - 2\right) * 1 + i \left((-2) * 3 + \frac{1}{2} * 1\right) = \left(\frac{3}{2} + 2\right) + i \left(-6 + \frac{1}{2}\right) \\ = \frac{7}{2} - i \frac{11}{2}; \end{array}$$

Let $z_{3} = z_{1}z_{2}$;

$$|z_{3}| = \sqrt{\left(\frac{7}{2}\right)^{2} + \left(\frac{-11}{2}\right)^{2}} = \sqrt{\frac{49}{4} + \frac{121}{4}} = \sqrt{\frac{170}{4}} = \frac{\sqrt{170}}{2} \approx 6.5192;$$ $$\varphi = \arcsin \left(\frac{-\frac{11}{2}}{\frac{7}{2}}\right) = \arcsin \left(-\frac{11}{7}\right);$$

Polar and exponential representation of $z_{1}z_{2}$:

$$z_{1} z_{2} = \frac{\sqrt{170}}{2} \left(\cos \left(\arcsin \left(-\frac{11}{7}\right)\right) + \sin \left(\arcsin \left(-\frac{11}{7}\right)\right)\right);$$ $$z_{1} z_{2} = \frac{\sqrt{170}}{2} e^{i \arcsin \left(-\frac{11}{7}\right)}.$$