Question

Find the modulus and the amplitude of the complex number

\frac{1 - 3 i}{1 + 2 i}

.

Accepted Answer

Let $z = \frac{1 - 3 i}{1 + 2 i}$

$∴ |z| = |\frac{1 - 3 i}{1 + 2 i}|$ (Apply modulus on both sides)

$∴ |z| = \frac{|1 - 3 i|}{|1 + 2 i|} [∵ |\frac{z_{1}}{z_{2}}| = \frac{|z_{1}|}{|z_{2}|}]$

$∴ |z| = \frac{\sqrt{1 + 9}}{\sqrt{1 + 4}} [∵ |x + i y| = \sqrt{x^{2} + y^{2}}]$

$∴ |z| = \frac{\sqrt{10}}{\sqrt{5}}$

$∴ |z| = \sqrt{2}$

Now, $z = \frac{1 - 3 i}{1 + 2 i}$

$\Rightarrow \arg (z) = \arg (\frac{1 - 3 i}{1 + 2 i})$ (Apply argument on both sides)

$\Rightarrow \arg (z) = \arg (1 - 3 i) - \arg (1 + 2 i) [∵ \arg (\frac{z_{1}}{z_{2}}) = \arg z_{1} - \arg z_{2}]$

Since, coordinates of numerator lies in the fourth quadrant and coordinates of denominator lies in the first quadrant.

$\Rightarrow \arg (z) = - \tan^{- 1} (\frac{3}{1}) - \tan^{- 1} (\frac{2}{1})$

$\Rightarrow \arg (z) = - \tan^{- 1} (3) - \tan^{- 1} (2)$

$\Rightarrow \arg (z) = - (\tan^{- 1} (3) + \tan^{- 1} (2))$

$\Rightarrow \arg (z) = - (π + \tan^{- 1} (\frac{3 + 2}{1 - 3.2})) [∵ \tan^{- 1} x + \tan^{- 1} y = π + \tan^{- 1} (\frac{x + y}{1 - x y}), x y > 1]$

$\Rightarrow \arg (z) = - (π + \tan^{- 1} (\frac{5}{1 - 6}))$

$\Rightarrow \arg (z) = - (π + \tan^{- 1} (- 1))$

$\Rightarrow \arg (z) = - (π - \frac{π}{4})$

Hence, the amplitude of the given complex number is $- \frac{3 π}{4}$ .

M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 3: EXERCISE 5.3

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 3: EXERCISE 5.3

Questions from M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 3: EXERCISE 5.3 with Hints & Solutions

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