The analytic function fz-z-1z2-1 has singularities at

F(z)=z 1z^2+1 Singularities are given by z^2+1=0 ⇒ z^2= 1 ⇒ z=±√( 1)=± i ...

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Standard XII

Mathematics

Equation of Conics in Complex Form

The analytic ...

Question

The analytic function $f (z) = \frac{z - 1}{z^{2} + 1}$ has singularities at

1

and

- 1

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1

and

1

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1

and

- i

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i

and

- i

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Solution

The correct option is D $i$ and $- i$
$f (z) = \frac{z - 1}{z^{2} + 1}$
Singularities are given by
$z^{2} + 1 = 0$
$\Rightarrow z^{2} = - 1$
$\Rightarrow z = \pm \sqrt{- 1} = \pm i$

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The analytic function f(z)=z−1z2+1 has singularities at

The correct option is D i and −if(z)=z−1z2+1 Singularities are given by z2+1=0 ⇒z2=−1 ⇒z=±√−1=±i

The analytic function $f (z) = \frac{z - 1}{z^{2} + 1}$ has singularities at

The correct option is D $i$ and $- i$
$f (z) = \frac{z - 1}{z^{2} + 1}$
Singularities are given by
$z^{2} + 1 = 0$
$\Rightarrow z^{2} = - 1$
$\Rightarrow z = \pm \sqrt{- 1} = \pm i$