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Show that the function f(z)=|z|^2 for z=x+iy is not differemti

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Show that the function $f (z) = ∣ z ∣^{2}$ for $z = x + i y$ is not differemtiable for $z \in C - {0}$ .

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Show that the function

f (z) = ∣ z ∣^{2}

for

z = x + i y

is not differemtiable for

z \in C - {0}

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

Given the function

f (z) = ∣ z ∣^{2}

or,

f (z) = x^{2} + y^{2} = u (x, y) + i v (x, y)

(Let).

Then

u = x^{2} + y^{2}

and

v = 0

Now,

u_{x} = 2 x, u_{y} = 2 y, v_{x} = 0

and

v_{y} = 0

So we have

u_{x} \neq = v_{y}

and

u_{y} \neq = - v_{x}

for

(x, y) \neq = (0, 0)

That is

f (z)

doesn't satisfy Cauchy-Riemann equation for

z \in C - {0}

So the function is not differentiable for

z \in C - {0}

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Question Text	Show that the function $f (z) = ∣ z ∣^{2}$ for $z = x + i y$ is not differemtiable for $z \in C - {0}$ .
Answer Type	Text solution:1
Upvotes	150

Show that the function f(z)=∣z∣2 for z=x+iy is not differemtiable for z∈C−{0}.

Text solutionVerified

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Show that the function $f (z) = ∣ z ∣^{2}$ for $z = x + i y$ is not differemtiable for $z \in C - {0}$ .