Question

find the modulus of (1+2i) (3-4i) ÷ (4+3i) (2-3i)
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Answer 1

Answer:

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

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