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

The correct option is [ 1/1 4i 2/1+i ]= [ 3 4i/5+i ] = [1+i 2+8i/(1 4i)(1+i) ] [ 3 4i/5+i ] = [ 1+9i/1+i 4i 4i^2 ] [ 3 4i/5+i ] = [ 1+9i/5 3i ] [ 3 4i/5+i ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Algebra of Complex Numbers

Reduce [1/1-4...

Question

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

Open in App

Solution

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

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

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

$= \frac{- 3 + 4 i + 27 i - 36 i^{2}}{25 + 5 i - 15 i - 3 i^{2}} = \frac{33 + 31 i}{28 - 10 i} \times \frac{28 + 10 i}{28 + 10 i}$

$= \frac{924 + 330 i + 868 i + 310 i^{2}}{(28)^{2} - (10 i)^{2}}$

$= \frac{614 + 1198 i}{784 + 100} (∵ i^{2} = - 1)$

$= \frac{2 (307 + 599 i)}{884} = \frac{307 + 599 i}{442}$

Suggest Corrections

292

Similar questions

Q. Reduce

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

to the standard form.

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

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

Q. Solve

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

Q. Simplify

(2 + 8 i) (1 - 4 i) - (3 - 2 i) (6 + 4 i)

(Note

: i = \sqrt{- 1}

)

Q. If

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

, then

a^{2} + b^{2} =

Join BYJU'S Learning Program

Reduce [11−4i−21+i]=[3−4i5+i] to the standard form.

[11−4i−21+i]=[3−4i5+i]=[1+i−2+8i(1−4i)(1+i)][3−4i5+i] =[−1+9i1+i−4i−4i2][3−4i5+i] =[−1+9i5−3i][3−4i5+i] =−3+4i+27i−36i225+5i−15i−3i2=33+31i28−10i×28+10i28+10i =924+330i+868i+310i2(28)2−(10i)2 =614+1198i784+100 (∵ i2=−1) =2(307+599i)884=307+599i442

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