Question 1071888
{{{4sqrt(3)+4i=8(sqrt(3)/2+I(1/2))}}}
{{{cos(30^o)=sqrt(3)/2}}} and {{{sin(30^o)=1/2}}} , so
{{{4sqrt(3)+4i=8(cos(30^o)+i*sin(30^o))}}}
 
TIP:
If they give you a complex number with
a non-zero real part, and
a different non-zero imaginary part,
and the angle has to be a nice round number,
one part will be {{{sqrt(3)}}} or {{{-sqrt(3)}}}
times the other part,
Because the only exact angles with
non-zero, different, and simple-to-write
cosine and sine are {{{30^o}}} , {{{60^o}}} in quadrant 1,
and angles symmetrical to those in the other quadrants.
{{{cos(30^o)=sqrt(3)/2=sin(60^o)}}} and
{{{cos(60^o)=1/2=sin(30^o)}}} .
If both parts are positive, it is one of those angles.
If there are negative parts,
it is a symmetrical angle on another quadrant.
For example, {{{-1+sqrt(3)*i}}}
has a negative real part, and a positive imaginary part,
so it would be plotted in quadrant 2.
The real part has to be cosine times something,
so it is {{{(1/2)(-2)=-1}}}
The quadrant 1 angle with {{{1/2}}} for a cosine
is {{{60^o}}} .
To get the symmetrical angle from quadrant 2,
You subtract from {{{180^o}}} ,
to get {{{180^o-60^o=120^o}}} in quadrant 2.
So,
{{{-1+sqrt(3)*i=2(-1/2+(sqrt(3)/2)*i)=2(cos(120^o)+i*sin(120^o))}}} .
If both parts are negative,
the complex number would be plotted in quadrant 3,
where the angle would be {{{180^o}}} plus
the symmetrical reference angle from quadrant 1.
If the real part is positive,
and the imaginary part negative,
You are in quadrant 4,
and the angle is {{{360^o}}} minus
the quadrant 1 angle.
 
FORMULAS:
Any complex number,
 {{{a + i b}}} ,
with {{{a}}} , and {{{b}}} being real numbers,
can always be written as
 {{{r (cos(theta) + i sin(theta))}}}
The formulas for {{{r}}} and {{{theta}}} are:
{{{r = sqrt (a^2 + b^2)}}}
 {{{cos(theta) = a/r}}}
and 
 {{{sin(theta)  = b/r}}} .
The last 2 formulas allow you to find theta
(in degrees or radians) with any scientific calculator.
If the calculator gives you a negative angle, add {{{360^o}}} (or {{{2pi}}} when working in radians).
(For faster calculations, you can use
{{{tan(theta) = b/a}}} and the signs, positive or negative, of {{{a}}} and {{{b}}} ).