find the product and its conjugate-complex numbers

**kikitaiszcraQk** · August 7th 2009, 06:43 PM

Can someone explain how I would solve this?

Find the product of the number and its conjugate.

square root of negative 15.

**malaygoel** · August 7th 2009, 07:05 PM

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?

**Plato** · August 8th 2009, 03:23 AM

Originally Posted by kikitaiszcraQk

Find the product of the number and its conjugate.

$\left( {a + bi} \right)\overline {\left( {a + bi} \right)} = \left( {a + bi} \right)\left( {a - bi} \right) = \left( a \right)^2 - \left( {bi} \right)^2 = a^2 + b^2$

**kikitaiszcraQk** · August 11th 2009, 09:26 AM

I don't understand how you can get "a" and "bi" from a negative square root.

**kikitaiszcraQk** · August 11th 2009, 09:54 AM

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?

**skeeter** · August 11th 2009, 10:37 AM

Originally Posted by kikitaiszcraQk

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

correct ...

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

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