Question

complex numbers

Answer 1

Express the complex number in trigonometric form.

-3 + 3 square root of threei

Answer 2

you need two things to write it as
\[r(\cos(\theta)+i\sin(\theta))\] namely \(r\) and \(\theta\)

Answer 3

\(r\) is easiest, it is the absolute value
the absolute value of \(a+bi\) is \(\sqrt{a^2+b^2}\) just like pythagoras
in your case it is
\[\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2\]

Answer 4

ooh sorry i can't read
it is \(-3+3\sqrt3 i\) right?

Answer 5

indeed c:

Answer 6

in that case \[r=\sqrt{3^2+(3\sqrt 3)^2}=\sqrt{9+27}=\sqrt{36}=6\]

Answer 7

is it clear that \((3\sqrt3)^2=27\) ?

Answer 8

no... because it woulx have to be 3*3*3 which would be 18 and when you square that you get 324

Answer 9

\((3\sqrt3)^2=3\times \sqrt3\times 3\times \sqrt3=3\times 3\times \sqrt3\times \sqrt3=9\times 3=27\)

Answer 10

that part is confusing sometimes
that is why i asked

Answer 11

in any case \(3\times 3\times 3=27\) not \(18\) and you don't square again, that is the square

Answer 12

now we have to find \(\theta\) and we do that via
\[\cos(\theta)=\frac{a}{r}\] and \[\sin(\theta)=\frac{b}{r}\]
in your case you get
\[\cos(\theta)=\frac{-3}{6}=-\frac{1}{2}\] and
\[\sin(\theta)=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2}\]