Re: Simple questions..
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Re: Simple questions..
In conjunction with this letter, see the many postings on Rectangular/Polar
Coordinate Conversion on the TI site at www.ti.com/calc. Select the TI-92
discussion group.
On your question about cis(theta):
cis(theta) corresponds on the TI-92 to e^(theta[2nd][i]).
First do the following:
Set the complex mode to RECTANGULAR.
Set the angle mode to RADIAN, NOT degree.
Enter delvar @ where @ is the GREEK letter, theta.
(If you have something stored in @, put it elsewhere temporarily.)
Enter e^(@[2nd][i]) [ENTER]
Look at the result.
Now set the complex mode to POLAR and do the following:
Switch to complex mode of POLAR.
When working with polar COMPLEX calculations involving angles on the TI-92,
ALWAYS SET THE CALCULATOR TO RADIAN, NOT degree.
If you plan to enter angles in degrees, use the superscript "o" character
accessed by [2nd][d] or from the MATH Angle menu.
You will probably want to have a function like the following to see what
the result is in degrees:
angle(c001)*180/PI->dan(c001)
Store this function.
Now suppose we want to do a calculation in APPROX mode:
5*cis(30 degrees) * 10*cis(50 degrees)
Enter 5e^(30[2nd][d][2nd]i) * 10e^(50[2nd][d][2nd][i]) [ENTER]
You will see (in FIX 3 format): e^(1.396*i)*50.000 BUT this is in radians.
You want to see it in degrees, so
Enter dan([2nd][ANS]) [ENTER]
You will see: 80.000 so the answer is:
5<30 degrees * 10<50 degrees = 50<80 degrees
Now lets do a radian calculation in AUTO mode:
5*cis(PI/6) * 10*cis((5/18)*PI)
Enter 5e^(PI/6[2nd][i])*10e^((5/18)PI[2nd][i]) [ENTER]
You will get an expression, take a look at it, then enter [<>][ENTER]
and you will see:
e^(1.396*i)*50.000.
Remember whether you enter a complex number as
e^(@[2nd][i])*R
or as
a+b[2nd][i]
the calculator keeps track of the complex number in the latter
rectangular format.
In rectangular format, make sure you use parenthesis appropriately:
(3+4i)/(5+6i) = 39/61+2/61*i
(3+4i)/5+6i = 3/5+34/5*i
3+4i/5+6i = 3+34/5*i
Why these 3 examples?
Set complex mode to RECTANGULAR, set AUTO mode.
Enter (3+4[2nd][i])/(5+6[2nd][i]) [ENTER]
Enter 3+4[2nd][i] [ENTER] then /5+6[2nd][i] [ENTER]
Enter 3+4[2nd][i]/5+6[2nd][i] [ENTER]
PI is [2nd][^]
"<" is the angle symbol just like [2nd][f] on the keyboard.
[<>] is the GREEN diamond key.
"->" is the [STO>] key.
Hope that helps.
---
In article <Pine.OSF.3.96.980312194051.12071A-100000@student.dei.uc.pt>,
"Joao Diogo O. Ramos" <jramos@STUDENT.DEI.UC.PT> wrote:
>
> I own a 92-II.
>
> The first question I'd like to put to you guys is how can I get
> the 3rd root of 8 (for example.) If you don't understood what I've asked
> (because of my "tecnical" English), what I'm looking for is the reverse of
> 2^3=8. Is there any way of doing that without doing 2^(1/3)???
>
> Now what made me write this email... I'm studying Complex numbers
> in school and I've figured out how to find zeros, solve equations, that
> stuff that uses cSolve, cFactor and so on... But I still haven't figured
> out how to make calculations with complex number ih the "trigonometrical"
> (I don't know how do you call it) form. To help you guys: He have complex
> number like this: z=a+ib and the other way that can be
> z=|z|cos(x)+i.|z|sin(x). But this last form is equal to z=|z|cis(x). We
> use a lot this last form, so it would be great for me if there was a way
> of make calculations with complex numbers just like that... Something that
> does the same like "cis" operation.
> I've look in the manual but couldn't find out nothing that could
> help me...
>
> Thanks in Advance...
> x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x
>
> ****** ++++++ **** ++++++ JODI -> URL:student.dei.uc.pt/~jramos
> ****** ++++++ ***** ++++++ ---------------------------------------
> ** ++ ++ ** ** ++ - Joao Diogo de Oliveira e Ramos -
> ** ** ++ ++ ** ** ++ - Eng. Informatica -> FCTUC/U.C. -
> ****** ++++++ ****** ++++++ ---------------------------------------
> ****** ++++++ **** ++++++ "DEI till death" -> URL:www.dei.uc.pt
>
> ----------jramos@student.dei.uc.pt ** jramos@alumni.dee.uc.pt------------
> -------------------------------------------------------------------------
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