MTH403 ASSIGNMENT NO. 1 FALL 2022

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MTH403 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || CALCULUS AND ANALYTICAL GEOMETRY - II || BY VuTech

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Assignment # 01 MTH403 (Fall 2022)

Maximum Marks: 10 Due Date: November 25, 2022

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Question # 1:

Find real `x` and `y` , if `(x -iy)(3+5i)` is the conjugate of `-6-24i`

Solution:

Let

`z=-6-24i`

`\bar z = \overline(-2-24i}`

And

`\bar z = (x-iy)(3+5i)`

`-6+24i = ( x-iy)(3+5i)`

`(x-iy)(3+5i) = -6+24i`

`(x-iy) = \frac{-6+24i}{3+5i} \times \frac{3-5i}{3-5i}`

`\left(x - iy\right) = \frac{\left( - 6 + 24i \right)\left(3 - 5i\right)}{\left(3 + 5i\right)\left(3 - 5i\right)}`

`\left(x - iy\right) = \frac{- 6\left(3 - 5i\right) + 24i\left(3 - 5i\right)}{\left( 3 \right)^2 - \left( 5i \right)^2}`

`\left(x - iy\right) = \frac{ - 18 + 30i + 72i - 120i^2}{9 - 25i^2}`

`\left(x - iy\right) = \frac{ - 18 + 102i + 120}{9 + 25}`

`\left(x - iy\right) = \frac{102 + 102i}{34}`

`\left(x - iy\right) = \frac{102}{34} + \frac{102i}{34}`

`\left(x - iy \right) = 3 + 3i`

`x = 3 and - y = - 3`

`x = 3 and y = 3`

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Question # 2:

Simplify `\left( \sqrt 3 + 3i \right)^{31}` by using De-Moivre’s theorem.

Solution:

De-Moivre’s theorem

`z = r^n\left(\cos n\theta + i\sin n\theta \right)` ------- (1)

Let

`z = \left( \sqrt 3 + 3i \right)^{31}`

`n = 31`

`\| z| = r = \sqrt {\left( \sqrt 3 \right)^2 + \left( 3 \right)^2}`

`\ = \sqrt {3 + 9}`

`\ = \sqrt {12}`

`\arg z = \theta = \tan ^{ - 1}\left(\frac{y}{x} \right)`

`\= \tan ^{ - 1}\left(\frac{\sqrt 3 }{3} \right)`

`\= \tan ^{ - 1}\left(\frac{\sqrt 3}{\sqrt 3 \sqrt 3 } \right)`

`\= \tan ^{ - 1}\left( \frac{1}{\sqrt 3 } \right)`

`\= \frac{\pi }{6}`

Now

`\left( \sqrt 3 + 3i \right)^{31} = \sqrt {12} ^{31}\left( \cos 31\left( \frac{\pi }{6} \right) + i\sin 31\left( \frac{\pi }{6} \right) \right)`

`\left( \sqrt 3 + 3i \right)^{31}= \sqrt {12} ^{31}\left( \cos \frac{31\pi }{6} + i\sin \frac{31\pi }{6} \right)`

`\left( \sqrt 3 + 3i \right)^{31} = \sqrt {12} ^{31}\left( cis\frac{31\pi }{6} \right)`

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