Put the following in the form of A - iB - 3 - 2i2 - 3i1 - 2i2 - i

The correct option is C 6325 1625 i (3 2i)(2+3i)(1+2i)(2 i) 6+9i 4i+62 i+4i+2 12+5i4+3i (12+5i)(4 3i)4^2+3^2 48 36i+20i+1525 ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

Put the follo...

Question

Put the following in the form of A + iB :
$\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$

\frac{3}{4} + \frac{9}{4} i

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\frac{63}{25} - \frac{16}{25} i

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\frac{5}{4} + \frac{9}{4} i

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

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Solution

The correct option is C $\frac{63}{25} - \frac{16}{25} i$
$\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$
$⟹ \frac{6 + 9 i - 4 i + 6}{2 - i + 4 i + 2}$
$⟹ \frac{12 + 5 i}{4 + 3 i}$
$⟹ \frac{(12 + 5 i) (4 - 3 i)}{4^{2} + 3^{2}}$
$⟹ \frac{48 - 36 i + 20 i + 15}{25}$
$⟹ \frac{63}{25}$ $- i \frac{16}{25}$

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Similar questions

Q. Express the following in

a + i b

form

\frac{4 i^{8} - 3 i^{9} + 3}{3 i^{11} - 4 i^{10} - 2}

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. Prove that the value of the determinant

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} - 7 & 5 + 3 i & \frac{2}{3} - 4 i 5 - 3 i & 8 & 4 + 5 i \frac{2}{3} + 4 i & 4 - 5 i & 9 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣

is real.

Q. Prove that the value of the determinant

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - 7 & 5 + 3 i & \frac{2}{3} - 4 i 5 - 3 i & 8 & 4 + 5 i \frac{2}{3} + 4 i & 4 - 5 i & 9 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

is real.

Q. Assertion :

7 + 4 i > 5 + 3 i,

where

i = \sqrt{- 1}

Reason: 7 > 5 and 4 > 3

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Put the following in the form of A + iB :(3−2i)(2+3i)(1+2i)(2−i)

The correct option is C 6325−1625i(3−2i)(2+3i)(1+2i)(2−i)⟹6+9i−4i+62−i+4i+2⟹12+5i4+3i⟹(12+5i)(4−3i)42+32⟹48−36i+20i+1525⟹6325−i1625

Put the following in the form of A + iB :
$\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$