Question

If \[z, \omega \ne 0.\] & \[|z|= |\omega|.\]
and \[\arg.(z)+\arg.(\omega)=\pi \space \space then \space \space z=...?\]

Answer 1

@satellite73 Please help:)

Answer 2

seems like we have an infinite number of possibilities
am i missing something?

Answer 3

\[z=re^{i\theta}\]
\[\omega=re^{i(\pi-\theta)}\]

Answer 4

\[\omega=r(\cos(\theta)+i\sin(\theta))\]
\[z=r(\cos(\pi-\theta)+i\sin(\pi-\theta))\]
\[z=r(\sin(\theta)+i\cos(\theta))\] by the cofunction identity, so if
\[\omega=a+b\] then
\[z=b+ai\]

Answer 5

@maheshmeghwal9
\[z=\left| z \right| e^{i \arg(z)}=\left| \omega \right| e^{i (\pi-\arg(\omega))}=\left| \omega \right|e^{i \pi}e^{-i \arg(\omega)}=-\left| \omega \right|e^{-i \arg(\omega)}=-\bar{\omega}\]
since \[e^{i \pi}=-1\]

Answer 6

@mukushla I have a doubt.
How did u write \[\huge{-|\omega|e^{-i \arg(\omega)}=- \bar \omega.}\]

Answer 7

i mean why.

Answer 8

\[\left| \omega \right|e^{-i \arg ( \omega)}=\left| \omega \right| (\cos (\arg ( \omega))-i \sin (\arg ( \omega))=\bar{\omega} \\ since\\ \omega=\left| \omega \right| (\cos (\arg ( \omega))+i \sin (\arg ( \omega))\]

Answer 9

oh i see
thanx a lot
i gt it:)