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Square Root of -1

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Square Root of -1

 
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  #1
Nov 22nd, 2008
Hello,

I'm writing a a program in C++ that adds/subtracts complex numbers. I'm having trouble writing the square root of -1 or "i" in a complex number.

Complex number:
realPart + imaginaryPart * i (where i is /-1) ["square root of -1"]

so far I'm thinking the code is something like this, but don't know how to go abouts to writing the square root part.

C++ Syntax (Toggle Plain Text) 
  1. double Complex::plusEq( Complex secondComplex )
  2. {
  3. 	double totalReal = real + secondComplex.getReal();
  4. 	double totalImag = imaginary + secondComplex.getReal();
  5.  
  6. 	return totalReal + totalImag + // i; <--------- this part
  7. }

I can't figure out how to write the commented "i".
Help please! ^^
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Re: Square Root of -1

 
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  #2
Nov 22nd, 2008
The square root of 1 is 1, this should lead to the square root of -1 being -1, however, this is imaginary. Taking the square root of a negative number doesn't work.

i = -1

Hope this helps
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Re: Square Root of -1

 
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Nov 22nd, 2008
So you're saying it should be the same as:

C++ Syntax (Toggle Plain Text) 
  1. int i = -1;
  2.  
  3. totalReal + totalImag + i;

Seems strange how the square root of -1 = -1...
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Re: Square Root of -1

 
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  #4
Nov 22nd, 2008
Well,
the square root of 4 is 2 because 2*2 is 4. So the square root of x is a number which complies to this: y*y = x.
This works for every number that is positive, however, if you try to take the square root of -1, that means you want y

y*y = x
-1 * -1 = oops, no, it's not x!
it's 1 instead of -1. That means there is no square root of -1. We just say it is -1 because we want to be able to calculate complex numbers with it.

Maybe my previous explanation was a little short, I hope this helps feel free to ask questions
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Re: Square Root of -1

 
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Nov 22nd, 2008
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Originally Posted by brechtjah View Post
Well,
the square root of 4 is 2 because 2*2 is 4. So the square root of x is a number which complies to this: y*y = x.
This works for every number that is positive, however, if you try to take the square root of -1, that means you want y

y*y = x
-1 * -1 = oops, no, it's not x!
it's 1 instead of -1. That means there is no square root of -1. We just say it is -1 because we want to be able to calculate complex numbers with it.

Maybe my previous explanation was a little short, I hope this helps feel free to ask questions
Sounds complicated lol, but I sorta get what you mean. =) So assuming a number * itself is = x
so if
2 * 2 = 4 // square root of 4 is equal to 2
-1 * -1 = doesn't work cuz all square roots are positive

so does this mean if i use
C++ Syntax (Toggle Plain Text) 
  1. totalReal + totalImaginary * 1 // or don't even bother with the *1 since it does nothing
should give me the correct answer?
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Re: Square Root of -1

 
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  #6
Nov 22nd, 2008
* -1
The square root of -1 is -1 we say. So it would just make your number negative.
Maybe this documentation is helpful for you too
http://en.wikipedia.org/wiki/Complex_number
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Re: Square Root of -1

 
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Nov 22nd, 2008
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Originally Posted by brechtjah View Post
* -1
The square root of -1 is -1 we say. So it would just make your number negative.
Maybe this documentation is helpful for you too
http://en.wikipedia.org/wiki/Complex_number
I will take a look at wikipedia for a more mathematical answer. Its weird but my teacher's sample output has a different answer than mine even if it is negative lol. Guess theres something else wrong with my code then lol. Thank you for your help brechtjah.
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Re: Square Root of -1

 
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  #8
Nov 22nd, 2008
If you don't mind add your code and I'll try and take a look at it
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Re: Square Root of -1

 
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  #9
Nov 22nd, 2008
sqrt(-1) = 1 * i
Where i can't be defined. You can't calculate it.

So if you were to return it as a result, it'd have to be a string like this:
"<Real part> + <Imaginary part> * i", in this case it would leave "0 + 1 * i", or "1i" in short.

Perhaps useful to add a String returning it this way to your class, or have a function do it.
Last edited by BeyondTheEye; Nov 22nd, 2008 at 8:59 am.
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Re: Square Root of -1

 
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  #10
Nov 22nd, 2008
If you are making a complex class you should return a complex object, and it should not be the first operand. I mean given z and w two complex number, the instruction z + w should not modify z! Instead, it should be used in an expression like c = z + w , assigning the result of the sum to another complex number c.

The member you need in this class are just Real and Imaginary, that are two double variables holding real and imaginary part of the compex number itself. You don't need to bother about calculating "i" (in fact, you couldn't even if you needed to). You should just append "i" as a char next to the imaginary value when you print the complex number on screen.

Here are examples:
C++ Syntax (Toggle Plain Text) 
  1. Complex Complex::operator+(Complex op2) {
  2. 	Complex temp;
  3. 	temp.Real = Real + op2.Real;
  4. 	temp.Imaginary = Imaginary + op2.Imaginary;
  5. 	return temp;
  6. }

to overload the operator+, and
C++ Syntax (Toggle Plain Text) 
  1. std::ostream& operator<<(std::ostream& lhs, Complex& obj) {
  2. 	lhs << "(";
  3. 	if(obj.Real>=0) {
  4. 		lhs << "+" << obj.Real;
  5. 	}
  6. 	else {
  7. 		lhs << obj.Real;
  8. 	}
  9. 	if(obj.Imaginary>=0) {
  10. 		lhs << " +" << obj.Imaginary << "i";
  11. 	}
  12. 	else {
  13. 		lhs << " " << obj.Imaginary << "i";
  14. 	}
  15. 	lhs << ")";
  16. 	return lhs;
  17. }

for output purposes.

Last but not least: square root of -1 is not -1.
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