The function w-u-iv-12logx2-y2-i tan-1 yx is not analytic at the point

Given function is w=u+iv=12log(x^2+y^2)+i tan^ 1 ( yx ) This is log z We have w=log z, nbsp;which is obviously not defined at z=0≈ (0,0) So, w=log z is not a ...

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Standard XII

Mathematics

Graphical Interpretation of Continuity

The function ...

Question

The function $w = u + i v = \frac{1}{2} l o g (x^{2} + y^{2}) + i t a n^{- 1} (\frac{y}{x})$ is not analytic at the point

(0,0)

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(0,1)

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(1,0)

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(2,

α

)

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Solution

The correct option is A (0,0)
Given function is
$w = u + i v = \frac{1}{2} l o g (x^{2} + y^{2}) + i t a n^{- 1} (\frac{y}{x})$
This is $l o g z$
We have $w = l o g z$ , which is obviously not defined at $z = 0 \approx (0, 0)$
So, $w = l o g z$ is not analytic at $(0, 0)$

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The function w=u+iv=12log(x2+y2)+itan−1(yx) is not analytic at the point

The correct option is A (0,0)Given function is w=u+iv=12log(x2+y2)+itan−1(yx) This is logz We have w=logz, which is obviously not defined at z=0≈(0,0) So, w=logz is not analytic at (0,0)

The function $w = u + i v = \frac{1}{2} l o g (x^{2} + y^{2}) + i t a n^{- 1} (\frac{y}{x})$ is not analytic at the point

The correct option is A (0,0)
Given function is
$w = u + i v = \frac{1}{2} l o g (x^{2} + y^{2}) + i t a n^{- 1} (\frac{y}{x})$
This is $l o g z$
We have $w = l o g z$ , which is obviously not defined at $z = 0 \approx (0, 0)$
So, $w = l o g z$ is not analytic at $(0, 0)$