(B) Given that $\bar{z} = \frac{1}{z}$.Multiplying both sides by $z$,we get $z \cdot \bar{z} = 1$.We know that for any complex number $z$,$z \cdot \ba

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(B) Given that $\bar{z} = \frac{1}{z}$.Multiplying both sides by $z$,we get $z \cdot \bar{z} = 1$.We know that for any complex number $z$,$z \cdot \bar{z} = |z|^2$.Therefore,$|z|^2 = 1$.Since the modulus $|z|$ must be non-negative,we have $|z| = 1$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Conjugate and Modulus of complex numbers

Medium

For any complex number $z$,$\bar{z} = \frac{1}{z}$ if and only if

A
$z$ is a purely real number
B
$|z| = 1$
C
$z$ is a purely imaginary number
D
$z = 1$

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Class 11 Questions

Class 11

Mathematics Questions

Chapter

1 Complex Numbers and Quadratic Equations: Complex Numbers

Subtopic

Conjugate and Modulus of complex numbers

What is the modulus of the complex number $(1+2i)(-2+i)$?

EasyAP EAMCET 2020

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Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

Medium

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If $x-iy = \sqrt{\frac{a-ib}{c-id}}$,prove that $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.

Medium

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If $\omega$ is a complex number satisfying $\left| \omega + \frac{1}{\omega} \right| = 2$,then the maximum distance of $\omega$ from the origin is:

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

DifficultKCET 2014

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For any complex number $z$,$\bar{z} = \frac{1}{z}$ if and only if

Similar Questions

What is the modulus of the complex number $(1+2i)(-2+i)$?

Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

If $x-iy = \sqrt{\frac{a-ib}{c-id}}$,prove that $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.

If $\omega$ is a complex number satisfying $\left| \omega + \frac{1}{\omega} \right| = 2$,then the maximum distance of $\omega$ from the origin is:

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

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