Question

Obtain the polar and exponential representations of z1, z2, and z1z2 where z1=1/2 - 2i, and z2=3+i

Answer 1

Answer on Question #78433 – Math – Complex Analysis

Question

Obtain the polar and exponential representations of $z_1, z_2$, and $z_1z_2$ where $z_1 = 1/2 - 2i$, and $z_2 = 3 + i$.

Solution

$$z_1 z_2 = \left(\frac{1}{2} - 2i\right)(3 + i) = \frac{7}{2} - \frac{11}{2}i$$

Exponential form:

$z_1$:

$$|z_1| = \sqrt{0.5^2 + (-2)^2} = \frac{\sqrt{17}}{2}, \quad \operatorname{Arg} z_1 = \tan^{-1}\left(\frac{-2}{1/2}\right) = \tan^{-1}(-4)$$ $$z_1 = |z_1| e^{i \operatorname{Arg} z_1} = \frac{\sqrt{17}}{2} e^{i \tan^{-1}(-4)}$$

$z_2$:

$$|z_2| = \sqrt{3^2 + 1^2} = \sqrt{10}, \quad \operatorname{Arg} z_2 = \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(1/3)$$ $$z_2 = |z_2| e^{i \operatorname{Arg} z_2} = \sqrt{10} e^{i \tan^{-1}(1/3)}$$

$z_1 z_2$:

$$|z_1 z_2| = |z_1| |z_2| = \frac{\sqrt{170}}{2}, \quad \operatorname{Arg} z_1 z_2 = \tan^{-1}\left(\frac{-11/2}{7/2}\right) = \tan^{-1}(-11/7)$$ $$z_1 z_2 = |z_1 z_2| e^{i \operatorname{Arg} z_1 z_2} = \frac{\sqrt{170}}{2} e^{i \tan^{-1}(-11/7)}$$

Polar form:

Since $e^{i\theta} = \cos \theta + i \sin \theta$, we obtain

$$z_1 = \frac{\sqrt{17}}{2} (\cos(\tan^{-1}(-4)) + i \sin(\tan^{-1}(-4)))$$ $$z_2 = \sqrt{10} (\cos(\tan^{-1}(1/3)) + i \sin(\tan^{-1}(1/3)))$$ $$z_1 z_2 = \frac{\sqrt{170}}{2} (\cos(\tan^{-1}(-11/7)) + i \sin(\tan^{-1}(-11/7)))$$

Answer:

$$z _ {1} = \frac {\sqrt {1 7}}{2} e ^ {i \tan^ {- 1} (- 4)} = \frac {\sqrt {1 7}}{2} (\cos (\tan^ {- 1} (- 4)) + i \sin (\tan^ {- 1} (- 4))),$$ $$z _ {2} = \sqrt {1 0} e ^ {i \tan^ {- 1} (1 / 3)} = \sqrt {1 0} (\cos (\tan^ {- 1} (1 / 3)) + i \sin (\tan^ {- 1} (1 / 3))),$$ $$z _ {1} z _ {2} = \frac {\sqrt {1 7 0}}{2} e ^ {i \tan^ {- 1} (- 1 1 / 7)} = \frac {\sqrt {1 7 0}}{2} (\cos (\tan^ {- 1} (- 1 1 / 7)) + i \sin (\tan^ {- 1} (- 1 1 / 7))).$$

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