Find the conjugate of 3-2i2-3i1-2i2-i

( 3 2i #10;)( 2 + 3i)( 1 + 2i)( 2 i #10;) = 6 + 9i 4i + 62 i + 4i + 2 #10; #10; = 12 + 5i4 + #10;3i #10; #10; = ( 12 + 5i ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

Find the conj...

Question

Find the conjugate of $\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$

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Solution

$\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)} = \frac{6 + 9 i - 4 i + 6}{2 - i + 4 i + 2}$

$= \frac{12 + 5 i}{4 + 3 i}$

$= \frac{(12 + 5 i) (4 - 3 i)}{16 - 9}$

$= \frac{48 - 36 i + 20 i + 15}{5}$

$= \frac{63 - 16 i}{5}$

The conjugate is $\frac{63 + 16 i}{5}$

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Similar questions

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

Q. Put the following in the form of A + iB :

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

Find the conjuagates of the following complex numbers :

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

Q. Find the conjugates of the following complex numbers:
(i) 4 − 5 i
(ii)

\frac{1}{3 + 5 i}

(iii)

\frac{1}{1 + i}

(iv)

\frac{(3 - i)^{2}}{2 + i}

(v)

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

(vi)

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

Q. The conjugate complex number of

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

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Find the conjugate of (3−2i)(2+3i)(1+2i)(2−i)

(3−2i)(2+3i)(1+2i)(2−i)=6+9i−4i+62−i+4i+2 =12+5i4+3i =(12+5i)(4−3i)16−9 =48−36i+20i+155 =63−16i5 The conjugate is 63+16i5

Find the conjugate of $\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$

$\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)} = \frac{6 + 9 i - 4 i + 6}{2 - i + 4 i + 2}$

$= \frac{12 + 5 i}{4 + 3 i}$

$= \frac{(12 + 5 i) (4 - 3 i)}{16 - 9}$

$= \frac{48 - 36 i + 20 i + 15}{5}$

$= \frac{63 - 16 i}{5}$

The conjugate is $\frac{63 + 16 i}{5}$