Question

Describe the geometric,polar and exponential representations of (5i+2)^-1

Answer 1

Answer on Question #85980 – Math – Complex Analysis

Question

Describe the geometric, polar and exponential representations of

$$(5i + 2)^{-1}.$$

Solution

Let's write the general view of the complex number:

$$z = a + ib.$$

Then (1) will be submitted as (2):

$$z = (5i + 2)^{-1} = \frac{1}{5i+2} = \frac{5i-2}{(5i+2)(5i-2)} = \frac{5i-2}{25i^2 - 4} = \frac{5i-2}{-25 - 4} = \frac{5i-2}{-29} = \frac{2}{29} - \frac{5}{29}i.$$

a) the geometric representations :

b) the polar representations:

general view of the polar representations is:

$$z = r \cos \varphi + ir \sin \varphi, \text{ where } r = \sqrt{a^2 + b^2}, \cos \varphi = \frac{a}{r}, \sin \varphi = \frac{b}{r}, \text{ then}$$ $$r = \sqrt{\left(\frac{2}{29}\right)^2 + \left(-\frac{5}{29}\right)^2} = \frac{\sqrt{29}}{29}, \cos \varphi = \frac{\frac{2}{29}}{\frac{\sqrt{29}}{29}} = \frac{2}{\sqrt{29}}, \sin \varphi = \frac{\frac{5}{29}}{\frac{\sqrt{29}}{29}} = \frac{-5}{\sqrt{29}}, \tan \varphi = -2.5, \varphi \approx -68.2{}^\circ$$ $$z = \frac {2}{2 9} - \frac {5}{2 9} i \approx \frac {\sqrt {2 9}}{2 9} \cos (- 6 8. 2 {}^ {\circ}) + i \frac {\sqrt {2 9}}{2 9} \sin (- 6 8. 2 {}^ {\circ}) = \frac {\sqrt {2 9}}{2 9} (\cos (- 6 8. 2 {}^ {\circ}) + i \sin (- 6 8. 2 {}^ {\circ})).$$

c) the exponential representations:

general view of the exponential representations is:

$$z = r e ^ {i \varphi}, \text { then } z \approx \frac {\sqrt {2 9}}{2 9} e ^ {- 6 8. 2 {}^ {\circ} i}.$$

Answer:

a) the geometric representations of $(5i + 2)^{-1}$ :

b) the polar representations of $(5i + 2)^{-1}$ :

$$(5 i + 2) ^ {- 1} \approx \frac {\sqrt {2 9}}{2 9} (\cos (- 6 8. 2 {}^ {\circ}) - i \sin (- 6 8. 2 {}^ {\circ}));$$

c) the exponential representations of $(5i + 2)^{-1}$ :

$$(5 i + 2) ^ {- 1} \approx \frac {\sqrt {2 9}}{2 9} e ^ {- 6 8. 2 {}^ {\circ} i}.$$

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