Question

$(1 + 2i) {}^{2} \times (1 - i) \\ find \: modulus \: and \: amplitude$
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Answer 1

$\underline{\large{\sf Answer:}}$

we have given the complex number

(1 + 2i )² (1 - i)

Z is denoted as a complex number

So ,

z = (1 + 2i)² (1 - i)

z = (1 + 2 (1)(2i) +(2i)²) (1-i)

.........[since ,using (a + b)² identity ]

z = (1 + 4i + 4i² )(1 - i)

= (1 + 4i - 4)(1 - i)

........[since, i² = (√-1)² = -1]

= (-3 + 4i) (1 - i)

= (-3 + 4i + 3i - 4i²)

= (-3 + 7i + 4 )

= ( 1 + 7i)

we know, if the complex no is

z = (x + yi) then their modulus is

|z| = √(x² + y²) , here note it that |z| is always positive i.e non-negative

therefor,

For z = ( 1 + 7i) ,x = 1 and y = 7

∴ $\sf |z| =\sqrt{x^2+y^2}$

∴ $\sf |z| =\sqrt{(1)^2 + (7)^2}$

∴ $\sf |z| = \sqrt{1 + 49}$

$\implies \sf \sqrt{50}= 5\sqrt{2}$

Now, we have to find amplitude,

$\sf \theta =tan^{-1}(\frac{y}{x})$

$\implies \sf tan^{-1}(\frac{7}{1})$

$\implies \sf tan^{-1}(7)$

Therefore , modulus of given complex number is 5√2 and their amplitude is $\sf tan^{-1}(7)$