Question

complex numbers anyone?

Answer 1

\[z=(1+i \sqrt{3})^{-2}\]

how do i write z in modulus .... and argument form?

Answer 2

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.

Answer 3

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

Answer 4

I could be completely wrong though

Answer 5

Oh I see. Euler notation.

Answer 6

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

Answer 7

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

Answer 8

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
\[(1+\sqrt3i){^-2}=\frac14 < \frac {2\pi }3\]