Question

find the modulus of (1+i)(2+i)/3+i​

Answer 1

Step-by-step explanation:

z = (1 + 2i)/(1 - 3i)

= (1 + 2i)(1 +3i)/(1 - 3i)(1 + 3i)

= {1(1 + 3i)+2i(1 +3i)}/{1²-(3i)²}

= ( 1 +3i +2i + 6i²)/(1 + 9)

= (-5 + 5i)/10

= (-1/2) + (1/2)i

now, modulus of the complex number is |(-1/2) + (1/2)i|

= √{(-1/2)² + (1/2)²}

= √{1/4 + 1/4}

= 1/√2

now, tan∅ = |Im(z)/Re(z)|

= |(1/2)/(-1/2)| = 1

tan∅ = tanπ/4

∅ = π/4

∅ lies on 2nd quadrant so,

arg(z) = π - ∅

= π - π/4

= 3π/4

Answer 2

The modulus of (1+i)(2+i)/3+i is 1

Given : (1+i)(2+i)/3+i

To find : The modulus

Solution :

Step 1 of 2 :

Write down the given number

The given complex number is

$\displaystyle \sf{ \frac{(1 + i)(2 + i)}{3 + i} }$

Step 2 of 2 :

Find the modulus

The required modulus

$\displaystyle \sf{ = \bigg| \frac{(1 + i)(2 + i)}{3 + i}\bigg| }$

$\displaystyle \sf{ = \frac{ | (1 + i) || (2 + i) | }{| 3 + i | } }$

$\displaystyle \sf{ = \frac{ \sqrt{ {1}^{2} + {1}^{2} } \: \: \sqrt{ {2}^{2} + {1}^{2} } }{ \sqrt{{3}^{2} + {1}^{2} } } }$

$\displaystyle \sf{ = \frac{ \sqrt{ 1 + 1 } \: \: \sqrt{ 4 + 1} }{ \sqrt{9 + 1} } }$

$\displaystyle \sf{ = \frac{ \sqrt{2 } \: \: \sqrt{5} }{ \sqrt{10} } }$

$\displaystyle \sf{ = \frac{ \sqrt{2 \times 5} }{ \sqrt{10} } }$

$\displaystyle \sf{ = \frac{ \sqrt{10} }{ \sqrt{10} } }$

$\displaystyle \sf{ = 1 }$

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