Symbolic operations with complex functions: conjugate, Re and Im

[Sergey]

Hello!

I am struggling a bit with the symbolic representation of some complex problems. Please, take a look at the attached file. 

How can I properly define complex function on form z=r*e^(i*phi)? Should I write z(r,phi)=r*e^(i*phi)? If I do so, then the symbolic evaluation of complex potentials (Phi and Psi) doesn't get along. Also, I am trying to use Re and Im functions to define the real and Imaginary parts of "z" without any success. The same goes to the conjugate of "z": it should be z*=r*e^(-i*phi).

Hopefully somebody can help me out.

UPDATE:

I did some progress, but still can't define properly function "z" without Mathcad telling that some variable is undefined. Updated sheet is attached.

Best regards, Sergey

[ttokoro]

[Werner_E · Accepted solution]

Defining z as a function with the arguments r and phi would be the preferred way if you want it actually to be seen as a function, but it works both ways.

z as a function:

z as a variable:

[Werner_E]

Just an additional note: You don't need to provide a guess value if you evaluate the solve block (find) symbolically.

To the community: Since which version of Prime has it been possible to evaluate solve blocks symbolically again?
This feature of real Mathcad was broken when Prime entered the stage and I wasn't aware that this option had been reintroduced.
I just tried with Prime 6 and it does not work but I never had installed P7 and P8, so all I can say that it works in P9.

Inserting a region with a symbolic evaluation as a line in a program still does not work as it did in Mathcad, though.

[terryhendicott]

What’s New in PTC Mathcad Prime 9.0.0.0
In this document, find topics introducing the enhancements in this release. Enhancements are categorized by functional area.
•Application Enhancements
◦Text Styles
◦Gradient Operator
◦Internal Links
•Symbolic Engine Enhancements
◦Symbolic Solving of ODEs
◦Logarithmic Integral Functions
◦Elliptic Integral Functions
◦Symbolic Solve Block (find)
◦Symbolic Assumption on Function
◦Definite Integral with complex limits
◦Improvement of Calculus Operators
◦General keyword improvements
◦General function improvements
•Numeric Engine Enhancements
◦PDESolve in Solve Block
◦General function enhancements
•Usability Enhancements
◦Custom Color Picker
◦Go-to Page
◦Current Page Tooltip
◦Math region to Text region conversion

Cheers

[Werner_E]

Ah, thanks! So P9 is the first version where its possible to symbolically evaluate solve blocks.

I tend not to read the "Whats New" document as of the glacier speed of Primes development, but I was aware of the gradient operator and the symbolic solve of ODEs but I missed symbolic solve block evaluation.

Its funny that they really dare to mention the introduction of a custom color picker in the ninth version of a software in 2023.

[Sergey]

Nice that we find it out 🙂

[Sergey]

Thank you very much for help. 

I am in love with Mathcad.

[Sergey]

How would you performed the symbolic derivation of the function z? When z is defined as a function of r and phi, the derivative is zero. 

[ttokoro]

[Sergey]

Thank you @ttokoro !!!

[Werner_E]

---

@Sergey wrote:

How would you performed the symbolic derivation of the function z? When z is defined as a function of r and phi, the derivative is zero. 

---

The derivative is zero because Phi(z(,r,phi)) simplifies to an expression in r and phi - no z! So the derivative with respect to z is zero.

What you possibly have in mind is easier to accomplish if z is not defined as a function

but it can also be done with z as a function, but a bit awkward as it uses an auxiliary variable (I used z_) and the modifier "substitute":

[Sergey]

THank you.

I find the solution given by @ttokoro is most suitable for this case. I will go forward with it.

Anyway, I appreciate the feedback and your solution.

[Werner_E]

Additional remark:

Depending on your needs, using "substitute" might be something you can use even if z is not defined with respect to r and phi, neither as a variable nor as a function: