The correct option is A 1 Given, #160; 1+2i / 1 ( 1 i ) ^ 2 = 1+2i 1 ( 1+ i ^ 2 2i ) #160; = 1+2i 1 ( 1 1 2i ) #160; =1 Ans: ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

1+2i/1- 1-i 2

Question

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

1

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

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

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- 1

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Solution

The correct option is A $1$
Given, $\frac{1 + 2 i}{1 - {(1 - i)}^{2}}$
$= \frac{1 + 2 i}{1 - (1 + i^{2} - 2 i)}$
$= \frac{1 + 2 i}{1 - (1 - 1 - 2 i)}$
$= 1$
Ans: A

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Similar questions

Q. Show that

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

is real.

(\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})

Q. Simplify

(\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 :

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

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1+2i1−(1−i)2

The correct option is A 1Given, 1+2i1−(1−i)2=1+2i1−(1+i2−2i)=1+2i1−(1−1−2i)=1Ans: A

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

The correct option is A $1$
Given, $\frac{1 + 2 i}{1 - {(1 - i)}^{2}}$
$= \frac{1 + 2 i}{1 - (1 + i^{2} - 2 i)}$
$= \frac{1 + 2 i}{1 - (1 - 1 - 2 i)}$
$= 1$
Ans: A