Z is a complex number then- z- - z 2 - -1 - -2Re z - z 2 - is equal to

The correct option is B z+1z Let z=x+iy ⇒ z^2=(x+iy)^2=x^2 y^2+2xyi ⇒ |z^2|=√((x^2 y^2)^2+4x^2y^2) ⇒ |z^2|=√((x^2+y^2)^2) ⇒ |z^2| ...

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Standard XII

Mathematics

Modulus of a Complex Number

z is a comple...

Question

$z$ is a complex number then, $\frac{z [∣ ∣ z^{2} ∣ ∣ - 1] + 2 R e (z)}{| z^{2} |}$ is equal to

z - \frac{1}{z}

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z + \frac{1}{z}

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R e (z)

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none of these

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Solution

The correct option is B $z + \frac{1}{z}$
Let $z = x + i y$

$\Rightarrow z^{2} = (x + i y)^{2} = x^{2} - y^{2} + 2 x y i$

$\Rightarrow | z^{2} | = \sqrt{(x^{2} - y^{2})^{2} + 4 x^{2} y^{2}}$

$\Rightarrow | z^{2} | = \sqrt{(x^{2} + y^{2})^{2}}$

$\Rightarrow | z^{2} | = x^{2} + y^{2}$

Now, consider $\frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = \frac{(x + i y) [x^{2} + y^{2} - 1] + 2 x}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = \frac{x (x^{2} + y^{2} - 1) + 2 x + i y (x^{2} + y^{2} - 1)}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = \frac{x (x^{2} + y^{2}) - x + 2 x + i y (x^{2} + y^{2}) - i y}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = \frac{x (x^{2} + y^{2}) + x + i y (x^{2} + y^{2}) - i y}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = \frac{(x + i y) (x^{2} + y^{2}) + (x - i y)}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = (x + i y) + \frac{x - i y}{x^{2} + y^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = z + \frac{¯ z}{| z |^{2}}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = z + \frac{¯ z}{z ¯ z}$

$\Rightarrow \frac{z [| z^{2} | - 1] + 2 R e (z)}{| z^{2} |} = z + \frac{1}{z}$

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z is a complex number then, z[∣∣z2∣∣−1]+2Re(z)|z2| is equal to

$z$ is a complex number then, $\frac{z [∣ ∣ z^{2} ∣ ∣ - 1] + 2 R e (z)}{| z^{2} |}$ is equal to