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True/false regarding C-R equation

True/False Questions for Complex AnalysisCauchy Riemann to find differentiable functionchecking differentiability on a multivariate functionSatisfies CR-equations, but is not complex differentiable in 0Differentiability of a complex function at a discontinuous pointDisproving differentiability of a complex functionA true statement with a false contrapositive?Which one of the following are true?Differentiability in the complex plane and in $Bbb R^2$.which of the following statement is True

0

$begingroup$

Is the following statement is True /false ?

The function $f$ defined by $$f(z) = begincases frac Im(z^2)bar z textif zneq 0 \ 0 text if z=0 endcases$$

satisfies the cauchy reimann equation at origin and also $f(z)$ is differentiable at origin?

My attempt : i think it will be true because for cauchy reimann equation here $Re(z)=0 $ so $f(z)$ will become constant that is satisfied the C-R equation and for differentiability here its is already given that $f(z)=0$ so $f'(z) =0$

Is its true ?

Any hints/solution will be appreciated

thanks u

complex-analysis derivatives cauchy-riemann-equation

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edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

$endgroup$

add a comment | 

0

$begingroup$

Is the following statement is True /false ?

The function $f$ defined by $$f(z) = begincases frac Im(z^2)bar z textif zneq 0 \ 0 text if z=0 endcases$$

satisfies the cauchy reimann equation at origin and also $f(z)$ is differentiable at origin?

My attempt : i think it will be true because for cauchy reimann equation here $Re(z)=0 $ so $f(z)$ will become constant that is satisfied the C-R equation and for differentiability here its is already given that $f(z)=0$ so $f'(z) =0$

Is its true ?

Any hints/solution will be appreciated

thanks u

complex-analysis derivatives cauchy-riemann-equation

share|cite|improve this question

edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

$endgroup$

add a comment | 

$begingroup$

Is the following statement is True /false ?

The function $f$ defined by $$f(z) = begincases frac Im(z^2)bar z textif zneq 0 \ 0 text if z=0 endcases$$

satisfies the cauchy reimann equation at origin and also $f(z)$ is differentiable at origin?

My attempt : i think it will be true because for cauchy reimann equation here $Re(z)=0 $ so $f(z)$ will become constant that is satisfied the C-R equation and for differentiability here its is already given that $f(z)=0$ so $f'(z) =0$

Is its true ?

Any hints/solution will be appreciated

thanks u

complex-analysis derivatives cauchy-riemann-equation

share|cite|improve this question

edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

$endgroup$

share|cite|improve this question

edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

share|cite|improve this question

edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

edited yesterday

José Carlos Santos

166k22132235

edited yesterday

José Carlos Santos

166k22132235

asked 2 days ago

jasminejasmine

1,873418

asked 2 days ago

jasminejasmine

1,873418

1 Answer
1

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2

$begingroup$

Note that, if $x,yinmathbb R$,$$f(x+yi)=frac2xyx-yi=frac2x^2yx^2+y^2+frac2xy^2x^2+y^2i.$$So, $u(x,y)=dfrac2x^2yx^2+y^2$ and $v(x,y)=dfrac2xy^2x^2+y^2$. So, both $u$ and $v$ are the null function when $x=0$ or $y=0$ and therefore $$dfracpartial upartial x(0,0)=dfracpartial upartial y(0,0)=dfracpartial vpartial x(0,0)=dfracpartial vpartial y(0,0)=0.$$So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.

However,$$lim_zto0fracf(z)-f(0)z=lim_zto0fracoperatornameIm(z^2)lvert zrvert^2$$and this limit doesn't exist. For instance, see what happens if $z$ is of the form $x+xi$, with $xin(0,infty)$. So, $f$ is not differentiable at $0$.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

add a comment | 

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2

$begingroup$

Note that, if $x,yinmathbb R$,$$f(x+yi)=frac2xyx-yi=frac2x^2yx^2+y^2+frac2xy^2x^2+y^2i.$$So, $u(x,y)=dfrac2x^2yx^2+y^2$ and $v(x,y)=dfrac2xy^2x^2+y^2$. So, both $u$ and $v$ are the null function when $x=0$ or $y=0$ and therefore $$dfracpartial upartial x(0,0)=dfracpartial upartial y(0,0)=dfracpartial vpartial x(0,0)=dfracpartial vpartial y(0,0)=0.$$So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.

However,$$lim_zto0fracf(z)-f(0)z=lim_zto0fracoperatornameIm(z^2)lvert zrvert^2$$and this limit doesn't exist. For instance, see what happens if $z$ is of the form $x+xi$, with $xin(0,infty)$. So, $f$ is not differentiable at $0$.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

add a comment | 

2

$begingroup$

Note that, if $x,yinmathbb R$,$$f(x+yi)=frac2xyx-yi=frac2x^2yx^2+y^2+frac2xy^2x^2+y^2i.$$So, $u(x,y)=dfrac2x^2yx^2+y^2$ and $v(x,y)=dfrac2xy^2x^2+y^2$. So, both $u$ and $v$ are the null function when $x=0$ or $y=0$ and therefore $$dfracpartial upartial x(0,0)=dfracpartial upartial y(0,0)=dfracpartial vpartial x(0,0)=dfracpartial vpartial y(0,0)=0.$$So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.

However,$$lim_zto0fracf(z)-f(0)z=lim_zto0fracoperatornameIm(z^2)lvert zrvert^2$$and this limit doesn't exist. For instance, see what happens if $z$ is of the form $x+xi$, with $xin(0,infty)$. So, $f$ is not differentiable at $0$.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

add a comment | 

$begingroup$

Note that, if $x,yinmathbb R$,$$f(x+yi)=frac2xyx-yi=frac2x^2yx^2+y^2+frac2xy^2x^2+y^2i.$$So, $u(x,y)=dfrac2x^2yx^2+y^2$ and $v(x,y)=dfrac2xy^2x^2+y^2$. So, both $u$ and $v$ are the null function when $x=0$ or $y=0$ and therefore $$dfracpartial upartial x(0,0)=dfracpartial upartial y(0,0)=dfracpartial vpartial x(0,0)=dfracpartial vpartial y(0,0)=0.$$So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.

However,$$lim_zto0fracf(z)-f(0)z=lim_zto0fracoperatornameIm(z^2)lvert zrvert^2$$and this limit doesn't exist. For instance, see what happens if $z$ is of the form $x+xi$, with $xin(0,infty)$. So, $f$ is not differentiable at $0$.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235

answered 2 days ago

José Carlos SantosJosé Carlos Santos

166k22132235