Reduce to the standard form ({1}{1-4i}-{2}{1+i})({3-4i}{5+i})

Question

Reduce to the standard form \(( \dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})\) to the standard form

Asked by Pragya Singh | 1 year ago | 50

0 0

Class 11 Maths Complex Numbers and Quadratic Equations Add Answer

Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number?

Solve the quadratic equations by factorization method only 6x² – 17ix – 12 = 0

Solve the quadratic equations by factorization method only x² + (1 – 2i)x – 2i = 0

Solve the quadratic equations by factorization method only x² + 10ix – 21 = 0

Solve the quadratic equations by factorization method only 17x² – 8x + 1 = 0

Accepted Answer

\( ( \dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})\)

\((\dfrac{1}{1-4i}-\dfrac{2}{1+i})[\dfrac{3-4i}{5+i}]\)

\([\dfrac{1+i-2+8i}{1+i-4i-4i^2}] [\dfrac{3-4i}{5+i}]\)

\([\dfrac{-1+9i}{5-3i}] [\dfrac{3-4i}{5+i}]\)

\([\dfrac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}]\)

\(\dfrac{33+31i}{2(14-5i)}\)

\( \dfrac{33+31i}{2(14-5i)}\times \dfrac{(14+5i)}{(14+5i)}\)

[On multiplying numerator and denominator by(14 + 5i)]

\( \dfrac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}\)

\( \dfrac{307+599i}{2(196-25i^2)}\)

\( \dfrac{307+599i}{2(221)}\) = \( \dfrac{307}{442}+\dfrac{599i}{442}\)

This is the required standard form

Answered by Abhisek | 1 year ago