1. ## hard complex question

captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
(b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.

2. Originally Posted by edgar davids
captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
(b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.

RonL

3. Originally Posted by edgar davids
captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
(b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.
Edgar has emailed me with the question he had meant to post when he
posted the above by mistake:

Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
equations to prove that in any of the following cases:
(a) Re(f) is constant;
(b) Im(f) is constant;
(c) |f| is constant;
one can conclude that f is constant.
Hint for part (c): calculate the partial derivatives of |f|2 = u2+v2 = const.

RonL

4. Originally Posted by CaptainBlack
Edgar has emailed me with the question he had meant to post when he
posted the above by mistake:

Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
equations to prove that in any of the following cases:
(a) Re(f) is constant;
(b) Im(f) is constant;
(a) Let $f(x,y) = u(x,y) + i v(x,y)$,

then the Cauchy-Riemann equations tell us that:

$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
$
,

and:

$
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
$

But we have $Re(f)$ is a constant, so:

$
\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}=0
$
,

and

$
\frac{\partial u}{\partial y} = 0
$

So along any curve such that $Re(z)=constant$; $f$ is a constant.

But a holomorphic function is determined every where by its values on a
curve, but $g(z)=Re(f(0))$ is holomorphic and equal to $f(z)$ on $Re(z)=0$,
so $f(z)=g(z)=Re(f(0))$ every where in $\mathbb{C}$.

Part (b) is similar.

RonL

5. ## thanks

how do you do part c? sorry to keep hassling you and really do appreciate all the help

all the best

6. Originally Posted by CaptainBlack

Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
equations to prove that in any of the following cases:
:
:
(c) |f| is constant;
one can conclude that f is constant.
Hint for part (c): calculate the partial derivatives of |f|2 = u2+v2 = const.
Now this last one is fiddly, and so prone to error, so you will need to
check this carefully.

$|f|$ constant implies $|f|^2$ is a constant, so:

$
\frac{\partial}{\partial x}|f|^2 = 2u \frac{\partial u}{\partial x} + 2 v\frac{\partial v}{\partial x}=0
$

and:

$
\frac{\partial}{\partial y}|f|^2 = 2u \frac{\partial u}{\partial y} + 2 v\frac{\partial v}{\partial y}=0
$

Hence we have:

$
u \frac{\partial u}{\partial x} = -v\frac{\partial v}{\partial x}
$
and $
u \frac{\partial u}{\partial y} = -v\frac{\partial v}{\partial y}
$
.

But $u$ and $v$ also satisfy the Cauchy-Riemann equations:

$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
$
and $
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
$
.

So with a bit of jiggery pokery we have:

$
u \frac{\partial u}{\partial x} = v\frac{\partial u}{\partial y}
$
and $
u \frac{\partial u}{\partial y} = -v\frac{\partial u}{\partial x}
$
.

Now multiply the first of these by $v$ and the second by $u$ to get:

$
vu \frac{\partial u}{\partial x} = v^2\frac{\partial u}{\partial y}
$
and $
u^2 \frac{\partial u}{\partial y} = -vu\frac{\partial u}{\partial x}
$
,

so:

$
u^2 \frac{\partial u}{\partial y} = -v^2\frac{\partial u}{\partial y}
$
,

Now chose a vertical line segment on which $f(z)$ has no zeros (such a segment exists because the zeros of a non-zero holomorphic function are isolated), then on this segment we have ${\partial u}/{\partial y} = 0$ (as if this were non zero at some point we would have $u^2+v^2=0$ which would mean that $f(z)=0$ at that point - a contradiction), so $u(z)$ is a constant on this segment.

Now I don't have the patience to do this myself, but I would expect a similar argument to show that on this segment ${\partial v}/{\partial y} = 0$, which implies that $v(z)$ is a constant on the segment, which with our previous result shows that $f(z)$ is constant on this segment, but a holomorphic function is determined everywhere by its values on a curve (of any kind) and so determined by its values on the segment. But on this segment $f(z)$ is equal to a constant function which is also holomorphic so $f(z)$ is constant on $\mathbb{C}$.

RonL