(z + a)(bar z + a) , where a is real, is equivalent to

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4-1.Complex numbers

easy

$(z + a)(\bar z + a)$, where $a$ is real, is equivalent to

$|z - a|$

${z^2} + {a^2}$

$|z + a{|^2}$

None of these

Solution

(c) As we know if $a$ is real, then $a = \overline a $
$(z + a)(\overline z + a) = (z + a)(\overline z + \overline a ) = (z + a)(\overline {z + a} )$ $ = |z + a{|^2}$

Standard 11

Mathematics

Which of the following are correct for any two complex numbers ${z_1}$ and ${z_2}$

normal

If $z_{1}=2-i, z_{2}=1+i,$ find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$

medium

Let $z,w$be complex numbers such that $\overline z + i\overline w = 0$and $arg\,\,zw = \pi $. Then arg z equals

medium

(AIEEE-2004)

If $|{z_1}|\, = \,|{z_2}|$ and $amp\,{z_1} + amp\,\,{z_2} = 0$, then

normal

If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

hard

(JEE MAIN-2023)