Arg Question (1 Viewer)

nrumble42 · Oct 30, 2016

Quick question guys:

Find arg (z) if z= (-2)/(1+i √2)
Give answer to 4dp.
Thanks!

InteGrand · Oct 30, 2016

nrumble42 said:
Quick question guys:

Find arg (z) if z= (-2)/(1+i √2)
Give answer to 4dp.
Thanks!

$\noindent Using $\frac{1}{w} = \frac{\overline{w}}{|w|^{2}}$, we have $z = -2\times \frac{1 - i\sqrt{2}}{1+2} = \frac{2}{3}\left(-1 + i \sqrt{2}\right)$. Hence $\mathrm{Arg}\left(z\right) = \tan^{-1} \left(\frac{\sqrt{2}}{-1}\right) + \pi = \pi - \tan^{-1} \left(\sqrt{2}\right)$. We can type this into a calculator to get it to $4$ decimal places (will be $2.1863$).$

si2136 · Oct 31, 2016

InteGrand said:
$\noindent Using $\frac{1}{w} = \frac{\overline{w}}{|w|^{2}}$, we have $z = -2\times \frac{1 - i\sqrt{2}}{1+2} = \frac{2}{3}\left(-1 + i \sqrt{2}\right)$. Hence $\mathrm{Arg}\left(z\right) = \tan^{-1} \left(\frac{\sqrt{2}}{-1}\right) + \pi = \pi - \tan^{-1} \left(\sqrt{2}\right)$. We can type this into a calculator to get it to $4$ decimal places (will be $2.1863$).$

In a test, can u just leave it as pi - tan inverse sqrt(2) if it doesn't ask for it in dp?

calamebe · Oct 31, 2016

si2136 said:
In a test, can u just leave it as pi - tan inverse sqrt(2) if it doesn't ask for it in dp?

Yes, I'd always leave it in the exact form unless it is specified not to.

Arg Question (1 Viewer)

nrumble42

Member

InteGrand

Well-Known Member

si2136

Well-Known Member

calamebe

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)