Question-126827

Question Number 126827 by harckinwunmy last updated on 24/Dec/20

Answered by AST last updated on 24/Dec/20

Q 33 .

\frac{4 + 7 i}{1 - 3 i}

Rationalising

\Rightarrow \frac{(4 + 7 i) (1 + 3 i)}{(1 - 3 i) (1 + 3 i)} = \frac{4 + 12 i + 7 i + (21 i^{2})}{1^{2} - {(3 i)}^{2}}

= \frac{4 + 19 i - 21}{1 - (- 9)} = \frac{- 17 + 19 i}{10} = \frac{- 17}{10} + i (\frac{19}{10})

= - 1.7 + 1.9 i (C)

Answered by AST last updated on 24/Dec/20

Q 34 .

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

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

= \frac{3 + i}{5} = \frac{3}{5} + \frac{1}{5} i

\Rightarrow x = \frac{3}{5}

Answered by AST last updated on 24/Dec/20

Q 35 .

Rationalising z

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

= \frac{1 - 2 i}{5} = \frac{1}{5} - \frac{2}{5} i

\bar{z} = \frac{1}{5} + \frac{2}{5} i

Answered by AST last updated on 24/Dec/20

Q 36 .

Note that \bar{z} = z

\Rightarrow conjugate of (conjugate of z) = z

Hence, answer is B .

Commented by harckinwunmy last updated on 24/Dec/20

t h a n k s A S T . c a n y o u h e l p c o m p l e t e i t

Answered by AST last updated on 24/Dec/20

Q 37 .

Rationalising z

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

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

\Rightarrow real part of z = \frac{1}{2}

Answered by AST last updated on 24/Dec/20

Q 38 .

4 + a - i = 2 b + a i

C omparing real and imaginary part

\Rightarrow a = - 1 and 4 + a = 2 b

\Rightarrow 4 - 1 = 2 b \Rightarrow b = \frac{3}{2}

Hence, (a, b) = (- 1, \frac{3}{2})

Answered by AST last updated on 24/Dec/20

Q 39 .

a + i b = {(2 - i)}^{2} = 4 + i^{2} - 4 i = 4 + (- 1) - 4 i

= 3 - 4 i

Comparing real and imaginary parts

a = 3, b = - 4

\Rightarrow a + b = 3 + (- 4) = - 1

Answered by AST last updated on 24/Dec/20

Q 40 .

Rationalising z

\Rightarrow \frac{1 (i)}{i (i)} = \frac{i}{i^{2}} = \frac{i}{- 1} = - i

Hence, \bar{z} = i

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