If a=|bar{a}| and b=|bar{b}|,then left(frac{b | vedclass

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

Question

(B) Given that $a = |\bar{a}|$ and $b = |\bar{b}|$. Since $|\bar{a}| = |a|$,we have $a = |a|$ and $b = |b|$.We know that for any complex number $z$,$z

Vedclass · Accepted Answer

$\left(\frac{\bar{a}-\bar{b}}{a b}\right)^2$

Vedclass · Answer

(B) Given that $a = |\bar{a}|$ and $b = |\bar{b}|$. Since $|\bar{a}| = |a|$,we have $a = |a|$ and $b = |b|$.We know that for any complex number $z$,$z \bar{z} = |z|^2$. Thus,$\bar{a} = \frac{|a|^2}{a} = \frac{a^2}{a} = a$ is not necessarily true,but rather $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{a \bar{a} \cdot a} = \frac{1}{a \bar{a}} \cdot \bar{a} = \frac{1}{a}$.Actually,using $z \bar{z} = |z|^2$,we have $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{|a|^2} = \frac{\bar{a}}{a \bar{a}} = \frac{1}{a}$.Similarly,$\frac{\bar{b}}{b^2} = \frac{1}{b}$.Therefore,$\left(\frac{\bar{a}}{a^2} - \frac{\bar{b}}{b^2}\right)^2 = \left(\frac{1}{a} - \frac{1}{b}\right)^2 = \left(\frac{b-a}{ab}\right)^2 = \left(\frac{a-b}{ab}\right)^2$.

If $a=|\bar{a}|$ and $b=|\bar{b}|$,then $\left(\frac{\bar{a}}{a^2}-\frac{\bar{b}}{b^2}\right)^2=$

Similar Questions

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$,then $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ is equal to

If the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$,then:

The modulus and amplitude of $\frac{1+2i}{1-(1-i)^{2}}$ are

If $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$,then $a^2 + b^2 = \dots$

Let $Z_1$ and $Z_2$ be any two complex numbers.
Statement $1: |Z_1 - Z_2| \ge |Z_1| - |Z_2|$
Statement $2: |Z_1 + Z_2| \le |Z_1| + |Z_2|$

Vedclass Test Series

Exam Paper Generator

Online Exam Module