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### #1 mortezahaydari

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Posted 30 April 2006 - 05:30 AM

you must use it numerical Re(2+ai)=2
but if i want to know Re((re(a)+i*re( c )) what should i do ,it must give {re(a)}

### #2 Overlord

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Posted 30 April 2006 - 11:16 AM

I don't understand,

if i enter re((re(a)+i*re( c )) it gives re(a) which is correct

re(2+a*i) is not 2, but 2-im(a) except if you have a â‚¬ IR : re(2+a*i)|im(a)=0 gives 2

### #3 vanhoa

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Posted 01 May 2006 - 07:18 AM

SO EASY
Varible a can have imagine number.
Re(2+ai)=/=2 => right because ai can have imagine number and then i miutiply whith that equa to a real number.

### #4 mortezahaydari

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Posted 01 May 2006 - 08:17 AM

SO EASY
Varible a can have imagine number.
Re(2+ai)=/=2 => right because ai can have imagine number and then i miutiply whith that equa to a real number.

what about Re( 1/ (Re( x ) + i* Re(y) ) )??????????????? no answer

### #5 vanhoa

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Posted 01 May 2006 - 09:42 AM

And what it give for that example?
It should be 1/re{a} because a might not be real only ( and "1/" is nothing ).
Is there any example more.

And in my afx when i solve the equation Abs(x)+Abs(x+1)=1, it got a result like this
X=0
X=1
X=X

### #6 PAP

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Posted 02 May 2006 - 01:44 PM

what about Re( 1/ (Re( x ) + i* Re(y) ) )??????????????? no answer

No answer, because x might be imaginary as well, or it might be equal to zero .

### #7 mortezahaydari

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Posted 03 May 2006 - 05:49 AM

No answer, because x might be imaginary as well, or it might be equal to zero .

no matter if x has imaganary or not ,Re(x) is real and re(y) is real

and if x=0 then answer is zero

### #8 vanhoa

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Posted 03 May 2006 - 07:24 AM

Sorry, I can not understand why there should be no answer.
The right must be 1/re(x), this because Re(x) is not equa to x when x is imagine.
i*Re(y) must be an imagine because Re(y) is real.

### #9 mortezahaydari

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Posted 06 May 2006 - 07:28 AM

Sorry, I can not understand why there should be no answer.
The right must be 1/re(x), this because Re(x) is not equa to x when x is imagine.
i*Re(y) must be an imagine because Re(y) is real.

i mean cp300 gives no answer

### #10 vanhoa

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Posted 06 May 2006 - 12:55 PM

Which mode did you use : complex mode or real mode?
It might make a big different!

### #11 Overlord

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Posted 06 May 2006 - 01:11 PM

and if x=0 then answer is zero

I think that's the problem, you have 2 different answers depending on the value of x and as there is no information on x, you can't give an unique answer because it can be wrong (and cp can't give multiple answers with conditions like "if x=0 then")

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Posted 06 May 2006 - 09:01 PM

I think that's the problem, you have 2 different answers depending on the value of x and as there is no information on x, you can't give an unique answer because it can be wrong (and cp can't give multiple answers with conditions like "if x=0 then")

The ClassPad, and every other CAS makes an assumption that x!=0 for most simplifications. Every CAS will simplify x/x to 0. This is wrong is x=0. If x=0 it is not defined.

### #13 mortezahaydari

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Posted 07 May 2006 - 06:58 AM

The ClassPad, and every other CAS makes an assumption that x!=0 for most simplifications. Every CAS will simplify x/x to 0. This is wrong is x=0. If x=0 it is not defined.

make me an example that Re( 1/ (Re( x ) + i* Re(y) ) ) will not be equal to Re(x)/((Re(x)^2+Re(y)^2)^(1/2)

also if x-->inf there should be inf

### #14 Guest_Guest_*

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Posted 07 May 2006 - 04:05 PM

make me an example that Re( 1/ (Re( x ) + i* Re(y) ) ) will not be equal to Re(x)/((Re(x)^2+Re(y)^2)^(1/2)

also if x-->inf there should be inf

Have to tried cexpand(Re( 1/ (Re( x ) + i* Re(y) ) ))

### #15 mortezahaydari

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Posted 10 May 2006 - 09:11 AM

Have to tried cexpand(Re( 1/ (Re( x ) + i* Re(y) ) ))

it works but now... another problem

cexpand (Re( 1/(x+y*i))=x/(x^2+y^2) that is wrong because x may not be real

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