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TheWrathOfMath

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z1 and z2 are two complex numbers such that z1*z2 is a real and non-zero number.
Prove that z1= x* (z2)bar, where x=real number.
 
Let [imath]z_1 = a + bi[/imath] and [imath]z_2 = c + di[/imath]. Multiply [imath]z_1^* z_2[/imath]. What are the imaginary terms?

-Dan
 
TheWrathOfMath said:
z1 and z2 are two complex numbers such that z1*z2 is a real and non-zero number.
Prove that z1= x* (z2)bar, where x=real number.
Click to expand...
First lets us cleanup the notation: [imath]z_1=z=(a+bi)~\&~z_2=w=(c+di)[/imath]
Then [imath]z\cdot w=(ac-bd)+(ad+bc)i[/imath].
If that product is real non-zero then [imath](ad+bc)=0~\&~(ac-bd)\not=0[/imath]
Here is the part that makes no sense whatsoever. The conjugate is [imath]\overline{\,w\,}=c-di[/imath]
You say that you are asked to prove that for some real [imath]t[/imath], [imath]z=t\cdot\overline{\,w\,}[/imath].
But how would that work??
 
TheWrathOfMath said:
It means that z1*z2=ac-bd, and ad = -bc
(or c/d = −a/b).

I highly appreciate your assistance, by the way.
Click to expand...
Since you want to relate [imath]z_1 = a+bi[/imath] to [imath]\bar{z_2} = c-di[/imath], how about using, not c/d = -a/b, but c/a = -d/b?
 
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