Question 1039836
<pre><b>
The complex number {{{1/2+i*sqrt(3)/2}}} is represented by
the vector (arrow) from the origin to the point. Its length
is r and its angle with the right side of the x-axis is <font face="symbol">q</font>.

{{{(matrix(1,3,1/2,",",sqrt(3)/2))}}}

{{{drawing(300,300,-1.5,1.5,-1.5,1.5,
graph(300,300,-1.5,1.5,-1.5,1.5),
locate(.1,.17,theta),  red(arc(0,0,.6,-.6,0,60)),
locate(.6,1,(matrix(1,3,1/2,",",sqrt(3)/2))),
line(0,0,0.5,0.8660254), line(0.48954715,0.76657321,0.5,0.8660254), line(0.4190983,0.80724688,0.5,0.8660254),
locate(.2,.6,r)


)}}}

We draw a perpendicular (in green) from the point to the
x-axis:

{{{drawing(300,300,-1.5,1.5,-1.5,1.5,
graph(300,300,-1.5,1.5,-1.5,1.5),
locate(.1,.17,theta),  red(arc(0,0,.6,-.6,0,60)),
locate(.6,1,(matrix(1,3,1/2,",",sqrt(3)/2))),
line(0,0,0.5,0.8660254), line(0.48954715,0.76657321,0.5,0.8660254), line(0.4190983,0.80724688,0.5,0.8660254),
green(line(0.5,0.8660254,0.5,0)), locate(.2,.6,r)

)}}}

That makes a right triangle and we can use the
Pythagorean theorem:

{{{r^2=(1/2)^2+(sqrt(3)/2)^2}}}
{{{r^2=1/4+3/4}}}
{{{r^2=4/4}}}
{{{r^2=1}}}
{{{r=1}}}

And since we know that {{{SINE=OPP/HYP}}},

{{{sin(theta)=(sqrt(3)/2)/1=sqrt(3)/2}}}
{{{theta="60°"}}} 

So the trigonometric form of  {{{1/2+i*sqrt(3)/2}}}

is {{{r(cos(theta)+i*sin(theta)^"")}}} or

{{{1(cos("60°")+isin("60°")^"")}}}

Or if yout teacher wants you to use radians, then 

{{{"60°"=pi/3}}}

{{{1(cos(pi/3)+isin(pi/3)^"")}}}.

Edwin</pre></b>