Question

help with complex no.s problem:--

Answer 1

What's your problem?

Answer 2

\[\arg.z+\arg. z'=0\] where z' is z's conjugate.We have to prove it!

Answer 3

\[
z= re^{i t}\\
Arg(z) =t\\
z'=re^{-i t}\\
Arg(z') =-t\\

\]
You can now conclude.

Answer 4

but why did u write Arg.(z)=t

Answer 5

In the exponential representation of a complex number z
\[
z = |z| e^{i \text { Arg } z}
\]

Answer 6

If the polar form isn't working for you, try this:\[z=a+bi \Rightarrow \text{Arg}(z)=\arctan \left( \frac{b}{a} \right)\]\[z'=a-bi \Rightarrow \text{Arg}(z)=\arctan \left( -\frac{b}{a} \right)\]Now, note what happens here:\[\arctan \left( \frac{b}{a} \right)+\arctan \left( -\frac{b}{a} \right)=0\]Since the arctangent function is odd, what does this imply? This will nudge you in the right direction for your proof. Remember, for proofs, you cannot start with the above statement, as we don't know it's true yet. Start with obvious facts, and then derive the above statement. Can you take it from here?

Answer 7

ya k!

Answer 8

\[z = a + i b \\
Arg z = \tan^{-1}\left( \frac b a\right)\\
z' = a- ib\\
Arg z' = \tan^{-1}\left(- \frac b a\right)=-\tan^{-1}\left( \frac b a\right)\\= - Arg z

\]

Answer 9

Slight correction: \(z'=a-bi \Rightarrow \text{Arg}(z')=\arctan \left( -\frac{b}{a} \right)\)

Answer 10

K! thanx to all I have got it