hp35s solve equation complex format

[dalupus]

Is there any way to solve the following directly without splitting off the imaginary part and using the 2x2 solver?

(1i0)x + (-.5i0.86603)y = 0i30

Or even better in polar

(1t0)x + (1t120)y = 30t-90 where t is theta

I saw stefan's matrix multi tool but even that assumes you need 2 equations which you don't in this case. That one equation is sufficient.

I could code this up simple enough using kramers rule, just curious if the functionality was built in.

Is it possible to use the 2x2 and 3x3 solver in a program?

[Dieter]

(06-11-2016 11:37 PM)dalupus Wrote:  I saw stefan's matrix multi tool but even that assumes you need 2 equations which you don't in this case. That one equation is sufficient.

Maybe I didn't get you right, maybe my skills in complex math are too limited, but what would you say is the correct and unique solution here?

What I see is x + (–0,5+0,86603 i)y = 30 i, which simply leads to the relation x = 30 i + (0,5–0,86603 i)y.

(06-11-2016 11:37 PM)dalupus Wrote:  I could code this up simple enough using kramers rule, just curious if the functionality was built in.

The 2x2 and 3x3 solvers seem to use Cramer's rule, but they do not support complex values.

(06-11-2016 11:37 PM)dalupus Wrote:  Is it possible to use the 2x2 and 3x3 solver in a program?

No, the solver is part of the equation list and this cannot be accessed from a program.

Dieter

[dalupus]

(06-12-2016 12:19 PM)Dieter Wrote:  

(06-11-2016 11:37 PM)dalupus Wrote:  I saw stefan's matrix multi tool but even that assumes you need 2 equations which you don't in this case. That one equation is sufficient.

Maybe I didn't get you right, maybe my skills in complex math are too limited, but what would you say is the correct and unique solution here?

What I see is x + (–0,5+0,86603 i)y = 30 i, which simply leads to the relation x = 30 i + (0,5–0,86603 i)y.

Thanks for the responses. For solving these equations you can break the real and imaginary parts into 2 equations. (assuming x and y are real, sorry should have stipulated that)
So the real part ends up being x -.5y =0
And the imaginary part is 0x + .86603y = 30

That is a bummer about not being able to use the solvers in a program. Guess I'll just roll my own extracting the real and imaginary parts into Cramer's rule

All in all though I have to say I am very impressed with the 35s. Much more powerful than the other calculators permitted on the FE exam.

[Dieter]

(06-12-2016 12:53 PM)dalupus Wrote:  Thanks for the responses. For solving these equations you can break the real and imaginary parts into 2 equations.

Sure. But that sounds a bit different from "That one equation is sufficient". ;-)

(06-12-2016 12:53 PM)dalupus Wrote:  So the real part ends up being x -.5y =0
And the imaginary part is 0x + .86603y = 30

This of course can be handled by the 2x2 solver.

(06-12-2016 12:53 PM)dalupus Wrote:  That is a bummer about not being able to use the solvers in a program.

Of course you can use the 35s solver in a program – the manual has several examples for this. But you cannot use equations from the equation list there. Which is what you want to do here.

(06-12-2016 12:53 PM)dalupus Wrote:  Guess I'll just roll my own extracting the real and imaginary parts into Cramer's rule

Should be easy to do. Each determinant can be calculated in merely five steps. But you do not have to implement Cramer's rule – you have to provide the six values for the two equations and store them in registers A...F. Then use the "2x2 lin. solve" function from the equation list.

In this case A=1, B=-0,5, C=0, D=0, E=0,86603, F=30  =>  x=17,32  y=34,64.

So you only have to do the extraction. If you want to solve the general equation (r1 + c1i)X + (r2+c2i)Y = r3+c3i you simply store

r1 in A
r2 in B
r3 in C
c1 in D
c2 in E
c3 in F

and run the 2x2 lin. solver without any entry. Be sure not to have any complex numbers in A...K.

But why bother about a program? Simply run the 2x2 solver and enter the three real parts, then the three imaginary parts. Simple as that.

(06-12-2016 12:53 PM)dalupus Wrote:  All in all though I have to say I am very impressed with the 35s. Much more powerful than the other calculators permitted on the FE exam.

I use it since it was available in 2007, and it has become my everyday calculation tool.

Dieter

[dalupus]

I meant using the 2x2 or 3x3 solver in a program, not the solver in general

The reason I wanted to do it programmatically versus entering the components individually into the solver is I already have the components I need on the stack in complex form from previous calculations.

i.e. my stack is

1i0
-.5i.86603
0i30

I guess I could simply do as you suggest and do a program to extract the components into the appropriate variables then use the built in solver.

Thanks for the help.

[dalupus]

This solution worked perfectly.

The 35s's treating complex numbers as a single unit actually came in handy since I can enter the numbers as a mix of rectangular and polar coordinates and it takes care of the translation in the background.

The program extracts out the real and imaginary parts of registers X,Y and Z and pops them into the A-F variables as you suggested.

[Csaba Tizedes]

(06-11-2016 11:37 PM)dalupus Wrote:  Is there any way to solve the following
(1i0)x + (-.5i0.86603)y = 0i30

Hi, Dalupus,

x and y are real or complex variables?

Csaba

[dalupus]

(06-13-2016 07:58 AM)Csaba Tizedes Wrote:  

(06-11-2016 11:37 PM)dalupus Wrote:  Is there any way to solve the following
(1i0)x + (-.5i0.86603)y = 0i30

Hi, Dalupus,

x and y are real or complex variables?

Csaba

They are real, which is why it is solvable. I corrected myself in a later post.