# find the product and its conjugate-complex numbers

• Aug 7th 2009, 07:43 PM
kikitaiszcraQk
find the product and its conjugate-complex numbers
Can someone explain how I would solve this?

Find the product of the number and its conjugate.

square root of negative 15.
• Aug 7th 2009, 08:05 PM
malaygoel
The product of a complex number and its conjugate is equal to the square of the modulus of the complex number.

Here,
complex number is $\sqrt{-15}$ or $i\sqrt{15}$
What is the modulus of this number?
What is conjugate of this number?
• Aug 8th 2009, 04:23 AM
Plato
Quote:

Originally Posted by kikitaiszcraQk
Find the product of the number and its conjugate.

$\left( {a + bi} \right)\overline {\left( {a + bi} \right)} = \left( {a + bi} \right)\left( {a - bi} \right) = \left( a \right)^2 - \left( {bi} \right)^2 = a^2 + b^2$
• Aug 11th 2009, 10:26 AM
kikitaiszcraQk
I don't understand how you can get "a" and "bi" from a negative square root.
• Aug 11th 2009, 10:54 AM
kikitaiszcraQk
Quote:

Originally Posted by malaygoel
The product of a complex number and its conjugate is equal to the square of the modulus of the complex number.

Here,
complex number is $\sqrt{-15}$ or $i\sqrt{15}$
What is the modulus of this number?
What is conjugate of this number?

The modulus of this number would be 15 so the product is 15^2=225?
• Aug 11th 2009, 11:37 AM
skeeter
Quote:

Originally Posted by kikitaiszcraQk
The modulus of this number would be 15 so the product is 15^2=225?

correct ...

$(0 + i\sqrt{15})(0 - i\sqrt{15}) = -i^2(\sqrt{15})^2 = -(-1)(15) = 15$