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

Answers (1)

Given problem is
\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )
Now, we will reduce it into

\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right ) = \left ( \frac{(1+i)-2(1-4i)}{(1+i)(1-4i)} \right )\left ( \frac{3-4i}{5+i} \right )
                                                         =\left ( \frac{1+i-2+8i}{1-4i+i-4i^2} \right )\left ( \frac{3-4i}{5+i} \right )
                                                         =\left ( \frac{-1+9i}{1-3i-4(-1)} \right )\left ( \frac{3-4i}{5+i} \right )
                                                         =\left ( \frac{-1+9i}{5-3i} \right )\left ( \frac{3-4i}{5+i} \right )
                                                         =\left ( \frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2} \right )= \left ( \frac{-3+31i+36}{25-10i+3} \right )= \frac{33+31i}{28-10i}= \frac{33+31i}{2(14-5i)}

Now, multiply numerator an denominator by  (14+5i)
\Rightarrow \frac{33+31i}{2(14-5i)}\times \frac{14+5i}{14+5i}
\Rightarrow \frac{462+165i+434i+155i^2}{2(14^2-(5i)^2)}                                        (using \ (a-b)(a+b)=a^2-b^2)
\Rightarrow \frac{462+599i-155}{2(196-25i^2)}
\Rightarrow \frac{307+599i}{2(196+25)}= \frac{307+599i}{2\times 221}= \frac{307+599i}{442}= \frac{307}{442}+i\frac{599}{442}

Therefore, answer is   \frac{307}{442}+i\frac{599}{442}

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