(z + a)(bar z + a),where a is real,is equival | vedclass

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(C) Given that $a$ is a real number,we have $a = \bar{a}$.Now,consider the expression $(z + a)(\bar{z} + a)$.Since $a = \bar{a}$,we can write this as

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$|z + a|^2$

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(C) Given that $a$ is a real number,we have $a = \bar{a}$.Now,consider the expression $(z + a)(\bar{z} + a)$.Since $a = \bar{a}$,we can write this as $(z + a)(\bar{z} + \bar{a})$.By the property of complex conjugates,$\bar{z} + \bar{a} = \overline{z + a}$.Therefore,$(z + a)(\bar{z} + a) = (z + a)(\overline{z + a})$.Using the property $|w|^2 = w \cdot \bar{w}$,we get $(z + a)(\overline{z + a}) = |z + a|^2$.

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

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