Real Part of a Function

[cj]

I'm used to seeing a function like:

\textbf{f}=x+iy \text

where

i = \sqrt{-1}

and understanding the the real part is:

\text{Re[f]} = x = Acos\theta

What, though, is the real part of a function like, for example,

\textbf{f}=\sqrt{x+iy}

??

[selfAdjoint]

Expand in Taylor's series, paying attention to powers of i, and separate terms with i out to get two series for the real and imaginary parts. This only works where the TS converges absolutely. Where is that?

[robphy]

Re(f)=\frac{f+f^*}{2}
where f^* is the complex-conjugate of f.

So,
\begin{align*}
Re(\sqrt{x+iy})&=\frac{(\sqrt{x+iy})+(\sqrt{x+iy})^*}{2}\\
&=\frac{\sqrt{x+iy}+\sqrt{x-iy}}{2}
\end{align*}


Additionally, you could write x+iy in polar coordinates.
Then, do the above.

[uart]

Since it looks like you're assuming the x and y are real and using the notation x = A cos(theta), y= A sin(theta) then the answer is just,

sqrt(A) cos(theta / 2)

[HallsofIvy]

You need to be more careful in distinguishing between the "real and imaginary parts" of the variable z= x+ iy, and the "real and imaginary parts" of the function value. Most text books write f(z)= u+ iv so that u is the real part and v the imaginary part of f: is z= x+ iy then u(x,y) and v(x,y) are real valued functions of the real variables of x and y.

[cj]

Since it looks like you're assuming the x and y are real and using the notation x = A cos(theta), y= A sin(theta) then the answer is just,

sqrt(A) cos(theta / 2)

If I am (and I am) going to use polar forms, the
general function:

\textbf{f}=x+iy

has the magnitude:

|\textbf{f}|=\sqrt{x^2+y^2}

with

x = Acos\theta \text{ and } y = Asin\theta

\text{BUT, I have the form: } \textbf{f}=\sqrt{x+iy} \text{ , NOT }\textbf{f}=x+iy

So how do I make use of

|\textbf{f}|=\sqrt{x^2+y^2}

since I don't have a function in the form of
\textbf{f}=x+iy

??

[HallsofIvy]

Have you simply ignored what everyone has said?

"I don't have a function in the form of f= x+ iy" is just saying you don't have the identity function: f(z)= z.

You're not going to be able to do very much is f(z)= z is the only function you know how to work with!

Suppose f(z)= z2. Then, writing z= x+iy, f(z)= (x+ iy)2=
x2+ 2ixy+ y2(i2)= (x2- y2)+ (2xy)i.
The real part of f is x2- y2 and the imaginary part of f is 2xy.

f(z)= √(z) is a little harder just because it is harder to calculate the value. It would probably be best to write z in polar form:
z= x+iy= r(cos(θ)+ i sin(θ)). The f(z)= √(z)= r1/2(cos(θ/2)+ i sin(θ/2)). The real part of f is r1/2cos(θ/2) and the real part is r1/2sin(θ/2).

[cj]

Got it -- thanks!

Although I think you meant "imaginary" rather than "real"
in the following:



... and the real part is r1/2sin(θ/2).



Have you simply ignored what everyone has said?

"I don't have a function in the form of f= x+ iy" is just saying you don't have the identity function: f(z)= z.

You're not going to be able to do very much is f(z)= z is the only function you know how to work with!

Suppose f(z)= z2. Then, writing z= x+iy, f(z)= (x+ iy)2=
x2+ 2ixy+ y2(i2)= (x2- y2)+ (2xy)i.
The real part of f is x2- y2 and the imaginary part of f is 2xy.

f(z)= √(z) is a little harder just because it is harder to calculate the value. It would probably be best to write z in polar form:
z= x+iy= r(cos(θ)+ i sin(θ)). The f(z)= √(z)= r1/2(cos(θ/2)+ i sin(θ/2)). The real part of f is r1/2cos(θ/2) and the real part is r1/2sin(θ/2).