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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$|z + a|^2$

Vedclass · Answer

(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

Similar Questions

If $x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$,then $x^2+y^2=$

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

If $\frac{(p + i)^2}{2p - i} = \mu + i\lambda$,then $\mu^2 + \lambda^2$ is equal to

If $z = x + iy$ with $xy \neq 0$ satisfies the equation $z^2 + i\bar{z} = 0$,then $|z^2|$ is equal to:

Find the modulus and argument of the complex number $\frac{1+2i}{1-3i}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module