11 - 2i - 31 - i 3 - 4i2 - 4i -

The correct option is D 14 + 94i (11+2i+31+i)(3+4i2 4i) ⇒(1+i+3 6i1+i+2i+2)(3+4i2 4i) ⇒(4 5i3 i)(3+4i2 4i) ⇒12+16i 15i+206 12i 2i 4 ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

11 - 2i + 31 ...

Question

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

\frac{1}{2} + \frac{9}{2} i

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\frac{1}{2} - \frac{9}{2} i

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\frac{1}{4} - \frac{9}{4} i

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\frac{1}{4} + \frac{9}{4} i

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Solution

The correct option is D $\frac{1}{4} + \frac{9}{4} i$
$(\frac{1}{1 + 2 i} + \frac{3}{1 + i}) (\frac{3 + 4 i}{2 - 4 i})$
$\Rightarrow (\frac{1 + i + 3 - 6 i}{1 + i + 2 i + 2}) (\frac{3 + 4 i}{2 - 4 i})$
$\Rightarrow (\frac{4 - 5 i}{3 - i}) (\frac{3 + 4 i}{2 - 4 i})$
$\Rightarrow \frac{12 + 16 i - 15 i + 20}{6 - 12 i - 2 i - 4}$
$\Rightarrow \frac{32 + i}{2 - 14 i} \times \frac{2 + 14 i}{2 + 14 i}$
$\Rightarrow \frac{63 + 448 i + 2 i - 14}{200}$
$\Rightarrow \frac{5 / 0 + 45 / 0 i}{20 / 0}$
$\Rightarrow \frac{1}{4} + \frac{9 i}{4}$ .

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Similar questions

Q. Simplify

(\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})

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. Show that

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

is real.

Q. Simplify

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

(Note

: i = \sqrt{- 1}

)

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(11−2i+31+i)(3+4i2−4i)=

The correct option is D 14+94i(11+2i+31+i)(3+4i2−4i)⇒(1+i+3−6i1+i+2i+2)(3+4i2−4i)⇒(4−5i3−i)(3+4i2−4i)⇒12+16i−15i+206−12i−2i−4⇒32+i2−14i×2+14i2+14i⇒63+448i+2i−14200⇒5/0+45/0i20/0⇒14+9i4.

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