Reduce 1 1-4i - 2 1-i 3-4i 5-i to the standard form-

Given: ( 1 / 1 4i 2 / 1+i #160;) ( 3 4i / 5+i #160;) =[ ( 1+i ) 2( 1 4i ) #160;/( 1 4i ) ( 1+i ) #160; #160;] ( 3 4i / 5+i #16 ...

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Byju's Answer

Standard XI

Mathematics

Algebra of Complex Numbers

Reduce 1 1...

Question

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

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Solution

Given: $(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i})$
$= [\frac{(1 + i) - 2 (1 - 4 i)}{(1 - 4 i) (1 + i)}] (\frac{3 - 4 i}{5 + i})$
$= (\frac{1 + i - 2 + 8 i}{1 + i - 4 i - 4 i^{2}}) (\frac{3 - 4 i}{5 + i})$
$= (\frac{- 1 + 9 i}{5 - 3 i}) (\frac{3 - 4 i}{5 + i})$
$= (\frac{- 3 + 4 i + 27 i - 36 i^{2}}{25 + 5 i - 15 i - 3 i^{2}})$
$= \frac{33 + 31 i}{28 - 10 i} = \frac{33 + 31 i}{2 (14 - 5 i)}$
$= \frac{(33 + 31 i)}{2 (14 - 5 i)} \times \frac{(14 + 5 i)}{(14 + 5 i)}$ [ on multiplying numerator and denominator by ( $14 + 5 i$ ) ]
$= \frac{462 + 165 i + 434 i + 155 i^{2}}{2 [{(14)}^{2} - {(5 i)}^{2}]} = \frac{307 + 599 i}{2 (196 - 25 i^{2})}$
$= \frac{307 + 599 i}{2 (221)} = \frac{307 + 599 i}{442} = \frac{307}{442} + \frac{599 i}{442}$

which is the required standard form.

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Similar questions

Reduce $[\frac{1}{1 - 4 i} - \frac{2}{1 + i}] = [\frac{3 - 4 i}{5 + i}]$ to the standard form.

Q. Solve

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

Express the following complex numbers in the standard form a + i b :

$(i) (1 + i) (1 + 2 i) (i i) \frac{3 + 2 i}{- 2 + i} (i i i) \frac{1}{(2 + i)^{2}} (i v) \frac{1 - i}{1 + i} (v) \frac{(2 + i)^{3}}{2 + 3 i} (v i) \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} (v i i) \frac{2 + 3 i}{4 + 5 i} (v i i i) \frac{(1 - i)^{3}}{1 - i^{3}} (i x) (1 + 2 i)^{- 3} (x) \frac{3 - 4 i}{(4 - 2 i) (1 + i)} (x i) (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) (x i i) \frac{5 + \sqrt{2 i}}{1 - \sqrt{2 i}}$

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

Q. Solve:

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

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