# I'm trying to find all functions such that Re(f(z)) + Im(f(z)) = 0, or

Gold Member
I'm trying to find all functions such that Re(f(z)) + Im(f(z)) = 0, or in other words...

$$Re(f(z)) = \left(\frac{1}{2} + \frac{i}{2}\right) f(z)$$

Obviously f(z) = 0 fits this. But is there an analytic way to either prove that this is the only function, or find others?

Mark44
Mentor

I'm trying to find all functions such that Re(f(z)) + Im(f(z)) = 0, or in other words...

$$Re(f(z)) = \left(\frac{1}{2} + \frac{i}{2}\right) f(z)$$

Obviously f(z) = 0 fits this. But is there an analytic way to either prove that this is the only function, or find others?
There are lots of others; for example, f(z) = 1 - i. In fact, f(z) = a - bi works, provided that a + b = 0.

For f(z) = 1 - i,
Re(f(z)) = 1 and (1/2)(1 + i) f(z) = (1/2)(1 + i)(1 - i) = (1/2)(1 + 1) = 1

chiro

I'm trying to find all functions such that Re(f(z)) + Im(f(z)) = 0, or in other words...

$$Re(f(z)) = \left(\frac{1}{2} + \frac{i}{2}\right) f(z)$$

Obviously f(z) = 0 fits this. But is there an analytic way to either prove that this is the only function, or find others?
Just out of curiosity, why do you want to find such functions? The space of functions in the set would probably be huge.

HallsofIvy
Homework Helper

All of Mark44's examples were constant functions but, obviously, such things as
f(z)= Re(z)+ Re(z)i, f(z)= Im(z)+ Im(z)i, or, more generally, f(z)= F(Re(z))(1+ i) or f(z)= F(Im(z))(1+ i), where F is any real valued function of a real variable, would work.

I think the most general form would be f(z)= F(z)(1+ i) where F can be any real valued function of z.

Gold Member

There are lots of others; for example, f(z) = 1 - i. In fact, f(z) = a - bi works, provided that a + b = 0.

For f(z) = 1 - i,
Re(f(z)) = 1 and (1/2)(1 + i) f(z) = (1/2)(1 + i)(1 - i) = (1/2)(1 + 1) = 1
All of Mark44's examples were constant functions but, obviously, such things as
f(z)= Re(z)+ Re(z)i, f(z)= Im(z)+ Im(z)i, or, more generally, f(z)= F(Re(z))(1+ i) or f(z)= F(Im(z))(1+ i), where F is any real valued function of a real variable, would work.

I think the most general form would be f(z)= F(z)(1+ i) where F can be any real valued function of z.
Awesome, thanks. You guys are always a great help.

Just out of curiosity, why do you want to find such functions? The space of functions in the set would probably be huge.
I was just curious, I guess. I think of questions like this sometimes.

All of Mark44's examples were constant functions but, obviously, such things as
f(z)= Re(z)+ Re(z)i, f(z)= Im(z)+ Im(z)i, or, more generally, f(z)= F(Re(z))(1+ i) or f(z)= F(Im(z))(1+ i), where F is any real valued function of a real variable, would work.

I think the most general form would be f(z)= F(z)(1+ i) where F can be any real valued function of z.
Don't you mean f(z)=F(z)(1-i), since we want that Re(f(z))=-Im(f(z)). Or am I really saying stupid things right now??

Gold Member

I just had a totally unrelated question, and I didn't want to start a new thread for it...

Is it possible for us to decompose a function into two functions, one even and one odd, that are added together? And how would this be done?

Sorry if that isn't clear...

I just had a totally unrelated question, and I didn't want to start a new thread for it...

Is it possible for us to decompose a function into two functions, one even and one odd, that are added together? And how would this be done?

Sorry if that isn't clear...
Yes!!

$$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$

Remember this trick well, you will meet it again in different contexts!!

Gold Member

Yes!!

$$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$

Remember this trick well, you will meet it again in different contexts!!
Wow, this is useful. I just checked it with the exponential function, and found that what sinh(x) and cosh(x) actually are are the odd and even components of e^x... fascinating!

The same trick can also be used to prove the following:

Show that any matrix A can be written as A=B+C, where B is symmetric and C is anti-symmetric (i.e. $${}^tC=-C$$)

Indeed, decompose

$$A=\frac{A+{}^tA}{2}+\frac{A-{}^tA}{2}$$

Or, show that any complex matrix A can be written as A=B+iC, where B and C are hermitian.

As you see, the same trick works in a variety of situations