Question:

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

Updated On: Sep 20, 2024
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Solution and Explanation

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

\(=[\frac{1+i-2+8i}{1+i-4i-4i^2}][\frac{3-4i}{5-3i}]=[\frac{-1+9i}{5-3i}][\frac{3-4i}{5+i}]\)

\([\frac{-3+4i+27i-36^2}{25+5i-15i-3i^2}]=\frac{33+31i-4i}{28-10i}=\frac{33+31i}{14-5i}\)

\(=\frac{(33+3li)}{2(14-5i)}×\frac{(14+5i)}{(14-5i)} \)     \([on\,multiplaying\,numerator\,and\,denominator\,by\,(14+5i)]\)

\(\frac{462+165i+434i+155i^2}{2[(14)^2(5i)^2]}=\frac{307+599i}{2(196-25i^2)}\)

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

This is the required standard form.

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Concepts Used:

Complex Number

A Complex Number is written in the form

a + ib

where,

  • “a” is a real number
  • “b” is an imaginary number

The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.