Find conjugate of the complex number 3-i2-i

The correct option is B 1 i z= 3+i 2 i =( 3+i ) ( 2+i ) #160; ( 2 i ) ( 2+i ) #160; = 6+5i+ i ^ 2 2 ^ 2 i ^ 2 = 5+5i 5 z=1+i z̅ ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

Find conjugat...

Question

Find conjugate of the complex number $\frac{3 + i}{2 - i}$

1 + i

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1 - i

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\frac{(1 + i)}{5}

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\frac{(1 - i)}{5}

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Solution

The correct option is B $1 - i$
$z = \frac{3 + i}{2 - i} = \frac{(3 + i) (2 + i)}{(2 - i) (2 + i)} = \frac{6 + 5 i + i^{2}}{2^{2} - i^{2}} = \frac{5 + 5 i}{5}$
$z = 1 + i$
$¯ z = \frac{3 - i}{2 + i} = 1 - i$

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Similar questions

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)}

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 multiplicative inverse of the following complex numbers:
(i) 1 − i
(ii)

(1 + i \sqrt{3})^{2}

(iii) 4 − 3i
(iv)

\sqrt{5} + 3 i

Q. Express the following complex numbers in the standard form a + i b:
(i)

(1 + i) (1 + 2 i)

(ii)

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

(iii)

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

(iv)

\frac{1 - i}{1 + i}

(v)

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

(vi)

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

(vii)

\frac{2 + 3 i}{4 + 5 i}

(viii)

\frac{(1 - i)^{3}}{1 - i^{3}}

(ix)

(1 + 2 i)^{- 3}

(x)

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

(xi)

(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})

(xii)

\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}

Q. The conjugate of complex number

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

is .

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Find conjugate of the complex number 3+i2−i

The correct option is B 1−iz=3+i2−i=(3+i)(2+i)(2−i)(2+i)=6+5i+i222−i2=5+5i5z=1+i¯z=3−i2+i=1−i

Find conjugate of the complex number $\frac{3 + i}{2 - i}$

The correct option is B $1 - i$
$z = \frac{3 + i}{2 - i} = \frac{(3 + i) (2 + i)}{(2 - i) (2 + i)} = \frac{6 + 5 i + i^{2}}{2^{2} - i^{2}} = \frac{5 + 5 i}{5}$
$z = 1 + i$
$¯ z = \frac{3 - i}{2 + i} = 1 - i$