# Complex variables : f(z) = zIm(z^2)?

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• Dec 6th 2012, 07:00 PM
mous99
Complex variables : f(z) = zIm(z^2)?
Complex variables: f(z) = z Im(z^2)?

If f(z) = z Im(z^2), determine the points in the complex plane such that f is differentiable.
• Dec 6th 2012, 07:40 PM
chiro
Re: Complex variables : f(z) = zIm(z^2)?
Hey mous99.

What have you tried for this question?

(Hint: Something is differentiable if it satisfies the Cauchy-Riemann equations).
• Dec 19th 2012, 07:15 AM
mous99
Re: Complex variables : f(z) = zIm(z^2)?
Let z = x+ iy z^2 = (x^2-y^2) + 2ixy or,
im(z^2 = 2ixy F(z) = z (2ixy))

Hence F(z) is differentiable at all points in complex plain except at (x,y)=(0,0) that is origin.

Is this answer right or wrong?
• Dec 19th 2012, 07:36 AM
Plato
Re: Complex variables : f(z) = zIm(z^2)?
Quote:

Originally Posted by mous99
Let z = x+ iy z^2 = (x^2-y^2) + 2ixy or,
im(z^2 = 2ixy F(z) = z (2ixy))

You have a serious mistake.

$\displaystyle \text{Im}(z^2)=2xy$. NOT $\displaystyle 2xyi$.
• Dec 19th 2012, 07:59 AM
mous99
Re: Complex variables : f(z) = zIm(z^2)?
why {Im}(z^2)=2xy?
• Dec 19th 2012, 08:07 AM
Plato
Re: Complex variables : f(z) = zIm(z^2)?
Quote:

Originally Posted by mous99
why {Im}(z^2)=2xy?

If indeed you do not know that, then you have no business trying this question.

If $\displaystyle z=x+yi$ then $\displaystyle \text{Re}(z)=x~\&~\text{Im}(z)=y$.

EXAMPLE:
$\displaystyle \text{Re}(xy+x^2i)=xy~\&~\text{Im}(xy+x^2i)=x^2$.
• Dec 19th 2012, 08:19 AM
mous99
Re: Complex variables : f(z) = zIm(z^2)?
Let z = x+ yi
Im (z2) = Im (x+yi) ^2
= Im (x^2- y ^ 2 +2xyi)

then how to continue for Im (z2) = 2xy???
• Dec 19th 2012, 08:36 AM
Plato
Re: Complex variables : f(z) = zIm(z^2)?
Quote:

Originally Posted by mous99
Let z = x+ yi
then how to continue for Im (z2) = 2xy???

$\displaystyle z\text{Im}(z^2)=(2x^2y)+i(2xy^2)$
• Dec 19th 2012, 08:50 AM
mous99
Re: Complex variables : f(z) = zIm(z^2)?
yes, i have same answer with you.
but how to determine the points 2x^2 y + 2xy^2 i in the complex plane such that f is differentiable?
• Dec 19th 2012, 08:54 AM
Plato
Re: Complex variables : f(z) = zIm(z^2)?
Quote:

Originally Posted by mous99
yes, i have same answer with you.
but how to determine the points 2x^2 y + 2xy^2 i in the complex plane such that f is differentiable?

(Hint: Something is differentiable if it satisfies the Cauchy-Riemann equations).
• Dec 23rd 2012, 05:26 AM
mous99
Re: Complex variables : f(z) = zIm(z^2)?
f(z) = z Im (z^2) = (x+iy) (2xy) = 2(x^2)y + 2x(y^2)i

determine the points in the complex plane such that f is differentiable.

df/dx=2xy+2y(x+iy), so i*df/dx=2ixy+2iy(x+iy)

And df/dy=2ixy+3x(x+iy)

This tell us f(z) is differentiable when x=iy.

is it right?