complex num. (1 Viewer)

elliottjohnson · Feb 11, 2010

how do you write

(3+4i)^-2

in mod-arg and cartesian form??

Can't figure out what to do with the -2

from Patel 2nd edition 4D Q.8

ninetypercent · Feb 11, 2010

cartesian form
$\frac{1}{(3+4i)^2} = \frac{1}{9 + 16i^2 + 24i} = \frac{1}{-7 + 24i} = \frac{1}{-7 + 24i}.\frac{-7-24i}{-7-24i} = \frac{-7 - 24i}{625}$

mod-arg form.
$3 + 4i = 5(cis) tan^-1\frac{4}{3}$

$(3 + 4i)^-2= 5^-2(cis) tan^-1\frac{-8}{3} = \frac{1}{25}(cis) tan^-1\frac{-8}{3}$

i think..

MetroMattums · Feb 11, 2010

Yeah, that's correct ninetypercent. Geddit?

No, seriously.

gurmies · Feb 11, 2010

Cazic · Feb 11, 2010

$\begin{align*}(3+4i)^{-2} &= \left( \sqrt{3^2 + 4^2}e^{i\tan^{-1}(4/3)} \right)^{-2} \\ &= \frac{e^{-2i\tan^{-1}(4/3)}}{25} \\ \end{align*}$

Note: e^ix = cos(x) + isin(x) if you must.

complex num. (1 Viewer)

elliottjohnson

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ninetypercent

ninety ninety ninety

MetroMattums

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gurmies

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Cazic

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