Solve- 11 - 2i - 31 - i3 - 4i2 - 4i-

( 11 2i+31+i ) ( 3+4i2 4i ) = ( 1+i+3(1 2i)(1 2i) ( 3+4i2 4i ) ) = ( 1+i+3 6i1+i 2i 2i^2 ) ( 3+4i2 4i ) = ( 4 5i2 i ) ( 3+4i2 4i ) = ( ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

Solve: 11 -...

Question

Solve: $(\frac{1}{1 - 2 i} + \frac{3}{1 + i}) (\frac{3 + 4 i}{2 - 4 i})$ .

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Solution

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

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

$= (\frac{1 + i + 3 - 6 i}{1 + i - 2 i - 2 i^{2}}) (\frac{3 + 4 i}{2 - 4 i})$

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

$= (\frac{12 + 16 i - 15 i - 20 i^{2}}{4 - 8 i - 2 i + 4 i^{2}})$

$= \frac{i + 32}{- 10 i} = \frac{- 1}{10} \frac{- 32}{10}$

$= \frac{- 1}{10} + \frac{32 i^{2}}{10 i}$

$= \frac{- 1}{10} + \frac{32 i}{10} = \frac{32 i - 1}{10}$

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Similar questions

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

Q. Simplify

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

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 :

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

Q. Show that

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

is real.

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Solve: (11−2i+31+i)(3+4i2−4i).

(11−2i+31+i)(3+4i2−4i)=(1+i+3(1−2i)(1−2i)(3+4i2−4i))=(1+i+3−6i1+i−2i−2i2)(3+4i2−4i)=(4−5i2−i)(3+4i2−4i)=(12+16i−15i−20i24−8i−2i+4i2)=i+32−10i=−110−3210=−110+32i210i=−110+32i10=32i−110

Solve: $(\frac{1}{1 - 2 i} + \frac{3}{1 + i}) (\frac{3 + 4 i}{2 - 4 i})$ .