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Mathematics 25 Online
OpenStudy (anonymous):

complex numbers anyone?

OpenStudy (anonymous):

\[z=(1+i \sqrt{3})^{-2}\] how do i write z in modulus .... and argument form?

OpenStudy (anonymous):

I'm not familiar with those forms, but I would just use the negative exponent to drop it to the denominator.. Then multiply top/bottom by the square of the conjugate and simplify.

OpenStudy (anonymous):

the argument is 2pi/3 and mod= 1/4 so\[\frac14<2\pi /3\]

OpenStudy (anonymous):

I could be completely wrong though

OpenStudy (anonymous):

Oh I see. Euler notation.

OpenStudy (anonymous):

if z is a complex no. written by \[z=r <\theta \]then z^2 is given by \[z^2=r^2< 2 \theta\]

OpenStudy (anonymous):

no wonder i coulddn't work it out... i hadn't been taught that name as a notation either.

OpenStudy (anonymous):

here \[z=\frac1{1+\sqrt3i}\] so \[r=1/\sqrt{1+3}=1/2\]\[\theta= \tan^{-1}\frac{\sqrt3}1=\frac {\pi}3\] so you have z^2 and u nedd to make square of r and 2 theta \[(1+\sqrt3i){^-2}=\frac14 < \frac {2\pi }3\]

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