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

##### Solution :-

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

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

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

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

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

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

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

[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

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