complex number

**saha.subham** · May 29th 2011, 04:52 AM

for any two complex no. z1 and z2 prove that re(z1 z2) = re(z1) * re(z2) - im(z1) * im(z2)
my incomplete soln-
don't have any idea about it.

**Also sprach Zarathustra** · May 29th 2011, 05:02 AM

Originally Posted by saha.subham

for any two complex no. z1 and z2 prove that re(z1 z2) = re(z1) + re(z2) - im(z1) * im(z2)
my incomplete soln-
don't have any idea about it.

Start by saying that z1=a+ib and z2=c+id.

If z=x+iy then:
The real part of z is:
Re(z)=x

and the imaginary part is:
Im(z)=y

**Prove It** · May 29th 2011, 05:13 AM

$\displaystyle z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$

$\displaystyle \begin{align*}z_1z_2 &= (x_1 + iy_1)(x_2 + iy_2) \\ &= x_1x_2 + ix_1y_2 + ix_2y_1 + i^2y_1y_2 \\ &= x_1x_2 - y_1y_2 + i(x_1y_2 + x_2y_1)\end{align*}$

So what is the real part of $\displaystyle z_1z_2$ ?

**saha.subham** · May 29th 2011, 07:04 AM

Originally Posted by Prove It

$\displaystyle z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$

$\displaystyle \begin{align*}z_1z_2 &= (x_1 + iy_1)(x_2 + iy_2) \\ &= x_1x_2 + ix_1y_2 + ix_2y_1 + i^2y_1y_2 \\ &= x_1x_2 - y_1y_2 + i(x_1y_2 + x_2y_1)\end{align*}$

So what is the real part of $\displaystyle z_1z_2$ ?

ok i have edited the question after consulting with my friend, it seems i have to remove the last part from ur answer, can u please guide me?

**Prove It** · May 29th 2011, 07:11 AM

You can write any complex number as $\displaystyle Z = X + iY$ , where $\displaystyle X$ is the real part of $\displaystyle Z$ and $\displaystyle Y$ is the imaginary part of $\displaystyle Z$ .

You have $\displaystyle z_1z_2 = x_1x_2 - y_1y_2 + i(x_1y_2 + x_2y_1)$ . If you're going to rewrite this as $\displaystyle Z= X + iY$ , what is $\displaystyle X$ (the real part of $\displaystyle z_1z_2$ )?

**saha.subham** · May 29th 2011, 09:51 PM

Originally Posted by Prove It

You can write any complex number as $\displaystyle Z = X + iY$ , where $\displaystyle X$ is the real part of $\displaystyle Z$ and $\displaystyle Y$ is the imaginary part of $\displaystyle Z$ .

You have $\displaystyle z_1z_2 = x_1x_2 - y_1y_2 + i(x_1y_2 + x_2y_1)$ . If you're going to rewrite this as $\displaystyle Z= X + iY$ , what is $\displaystyle X$ (the real part of $\displaystyle z_1z_2$ )?

How thus is in the form
re(z1) * re(z2) - im(z1) * im(z2) ???

**Prove It** · May 29th 2011, 11:00 PM

Because $\displaystyle x_1$ is the real part of $\displaystyle z_1$ , $\displaystyle x_2$ is the real part of $\displaystyle z_2$ , $\displaystyle y_1$ is the imaginary part of $\displaystyle z_1$ and $\displaystyle y_2$ is the imaginary part of $\displaystyle z_2$ .

**saha.subham** · May 30th 2011, 03:54 AM

Originally Posted by Prove It

Because $\displaystyle x_1$ is the real part of $\displaystyle z_1$ , $\displaystyle x_2$ is the real part of $\displaystyle z_2$ , $\displaystyle y_1$ is the imaginary part of $\displaystyle z_1$ and $\displaystyle y_2$ is the imaginary part of $\displaystyle z_2$ .

then it will be,
x1x2 - y1y2,
but how can i cancel out i(x1y2+ x2y1)??

**Prove It** · May 30th 2011, 04:13 AM

You DON'T. You are asked merely to state the REAL part of $\displaystyle z_1z_2$ , in other words, everything that's not attached to an $\displaystyle i$ .

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