PDA

View Full Version : Separating real and complex parts of a number


KarolisK
Oct10-11, 02:12 AM
1. The problem statement, all variables and given/known data
Hello, I am supposed to express the and the phase part of expression:

\displaystyle{S=\frac{k}{\sqrt{1+i\gamma_0}} \cdot exp\left(\frac{z}{1 + i\gamma_0}\right)}

2. Relevant equations
The answer should be in the form:

\displaystyle{S=a(\gamma_0) \cdot exp\left(i\varphi(\gamma_0)\right)}


3. The attempt at a solution
Well, its clear(probably) that for the amplitude part I just have to multiply this equation by its complex conjugate and take a square root out of the result. This leaves me with the expression of:
\displaystyle{ a(\gamma_0)=\frac{k}{\left(1+\gamma^2_0\right)^{1/4}} \cdot exp\left(\frac{z}{1+\gamma^2_0}\right) }

However, I dont quite understand how to get the complex(phase) part of the number? A hint where to start would be very gladly accepted :), thank you.

ehild
Oct10-11, 02:48 AM
]
Well, its clear(probably) that for the real part I just have to multiply this equation by its complex conjugate and take a square root out of the result.

It is the magnitude or absolute value of the complex number instead of the real part. So how do you get the real and imaginary parts?


ehild

KarolisK
Oct10-11, 11:30 AM
Ah yes, sorry, its the amplitude, thanks for noticing. I'll reformulate the problem. I need to get the expressions for the amplitude and the phase.

ehild
Oct10-11, 11:59 AM
Is z in the exponent a complex number? Then write it out with its real and imaginary parts, and find the real and imaginary parts of z/(1+iγ0), so you have the exponential in the form exp(u+iv) .

Write the fraction 1/(1+iγ0) in exponential form, too: exp(a+ib). Then your formula is equal to k*exp((a+u)+i(b+v)). k*exp(a+b) is the magnitude, and the phase is b+v.

ehild

KarolisK
Oct11-11, 01:58 AM
z is real and negative and k is real and positive constant. Anyway, expressing the fraction sqrt(1/(1+iy0) can get me just as close as:

\displaystyle{ k \cdot exp \left(\frac{1}{2}ln \left( \frac{1}{1+\gamma^2_0}-\frac{i\gamma_0}{1+\gamma^2_0}\right) \right) }

Which I dont understand how to simplify to form exp(a+ib). I have also tried expressing phase from the general form:

\displaystyle{ \frac{S}{a(\gamma_0)}=exp \left(i \varphi \right) }

Which is kinnda closer to the answer with the expression:

\displaystyle{ i\varphi=\frac{1}{4}ln\left(\frac{1-i\gamma_0}{1+i\gamma_0} \right) -
\frac{i\gamma_0z}{1+\gamma^2_0} }

The answer should be:

\displaystyle{\varphi = -\frac{1}{2}arctan\ \gamma_0 - \frac{\gamma_0z}{1+\gamma^2_0} }

ehild
Oct11-11, 02:45 AM
Write all factors in the exponential form: r*eiφ. Any complex number u+iv= r*eiφ, where r=\sqrt{(u^2+v^2) } and tan(φ)=v/u.
This way, 1+i \gamma_0=\sqrt{1+\gamma_0 ^2} e^ {i\arctan(\gamma_0)}

ehild

KarolisK
Oct11-11, 11:45 AM
ah yes, thank you very much:)