Question 200450
<pre><font size = 4 color = "indigo"><b>
{{{i = 0+1i}}} so the number i is the vector that goes from the
origin to the point (0,1)
 
That vector has an angle of {{{pi/2}}} or 90° as we see by the
blue arrow.  I'll use this "at" symbol, " @ ", for theta.
 
So @ = {{{pi/2}}} and r = 1.
 
Trig form for a complex number is 
 
{{{r(cos("@")+i*sin("@"))}}}
 
{{{drawing(300,300,-2,2,-2,2, line(-.1,.7,0,1),line(.1,.7,0,1), 
rectangle(0,0,0,1), line(-.015,0,-.015,1), blue(arc(0,0,.5,-.5,0,90) ),
graph(300,300,-2,2,-2,2), triangle(0,0,0,1,0,0) )}}}
 
However {{{"@"}}} doesn't have to just be identified as {{{pi/2}}} because 
we may add any multiple of {{{2pi}}} to the angle and it will be in
the exact same position. So we use {{{"@"=pi/2+2npi}}} where {{{n}}}
represents any integer, positive negative or zero.  
 
So {{{r(cos("@")+i*sin("@"))}}}
 
becomes:
 
{{{r(cos("@"+2pi*n)+i*sin("@"+2pi*n))}}}
 
And for both problems:
 
{{{i=1(cos(pi/2+2pi*n)+i*sin(pi/2+2pi*n))=1(cos(pi(1/2+2n))+i*sin(pi(1/2+2n)))}}}=
 
{{{1(cos(pi((1+4n)/2))+i*sin(pi((1+4n)/2)))}}}
 
Now first we'll take the square root, we raise that to the {{{1/2}}}
power:
 
{{{(1(cos(pi((1+4n)/2))+i*sin(pi((1+4n)/2))))^(1/2)}}}
 
Now the rule for simplifying a complex number in trig
form is:
 
1. Raise the modulus (lenth of vector) to the power 
2. Multiply the argument (angle) by the power.
 
{{{1^(1/2)(cos((1/2)pi((1+4n)/2))+i*sin((1/2)pi((1+4n)/2)))}}}
 
which becomes:
 
{{{1(cos(pi((1+4n)/4))+i*sin(pi((1+4n)/4)))}}}
 
or
 
{{{cos(pi((1+4n)/4))+i*sin(pi((1+4n)/4))}}}
 
Now we let n take on any two consecutive integer values, but
the easiest are n=0 and n=1, so letting n=0:
 
{{{cos(pi((1+4*0)/4))+i*sin(pi((1+4*0)/4))}}}
 
{{{cos(pi/4)+i*sin(pi/4) = sqrt(2)/2+i*sqrt(2)/2}}}
 
That is this vector:
 
{{{drawing(300,300,-2,2,-2,2, line(sqrt(2)/2,sqrt(2)/2,0,0), 
blue(arc(0,0,.5,-.5,0,45) ),
graph(300,300,-2,2,-2,2),
line(sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2,sqrt(2)/2), 
line(sqrt(2)/2,sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2)
 

 )}}}
 
Now letting n=1:
 
{{{cos(pi((1+4*1)/4))+i*sin(pi((1+4*1)/4))}}}
 
{{{cos(5pi/4)+i*sin(5pi/4) = -sqrt(2)/2+i*(-sqrt(2)/2)= -sqrt(2)/2-i*sqrt(2)/2  }}}
 
That is this vector:
 
{{{drawing(300,300,-2,2,-2,2, line(-sqrt(2)/2,-sqrt(2)/2,0,0), 
blue(arc(0,0,.5,-.5,0,225) ),
graph(300,300,-2,2,-2,2),
line(-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2), 
line(-sqrt(2)/2,-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2)
 

 )}}}
 
So the two square roots of i are 
 
 
 
 
 

{{{sqrt(2)/2+i*sqrt(2)/2  }}} and {{{-sqrt(2)/2-i*sqrt(2)/2  }}}
 
Putting them both on the same graph:
 
{{{drawing(300,300,-2,2,-2,2, line(sqrt(2)/2,sqrt(2)/2,0,0), 
blue(arc(0,0,.5,-.5,0,45) ), line(-sqrt(2)/2,-sqrt(2)/2,0,0),
graph(300,300,-2,2,-2,2), blue(arc(0,0,.7,-.7,0,225) ), 
line(sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2,sqrt(2)/2), 
line(sqrt(2)/2,sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2),
line(-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2), 
line(-sqrt(2)/2,-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2)
 

 )}}}
 
They form two equally spaced "spokes of a wheel":
 
{{{drawing(300,300,-2,2,-2,2, line(sqrt(2)/2,sqrt(2)/2,0,0), 
blue(arc(0,0,.5,-.5,0,45) ), line(-sqrt(2)/2,-sqrt(2)/2,0,0),
graph(300,300,-2,2,-2,2), blue(arc(0,0,.7,-.7,0,225) ), 
line(sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2,sqrt(2)/2), 
line(sqrt(2)/2,sqrt(2)/2-.1,sqrt(2)/2,sqrt(2)/2),
line(-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2), 
line(-sqrt(2)/2,-sqrt(2)/2+.1,-sqrt(2)/2,-sqrt(2)/2),
circle(0,0,1)
 

 )}}}
 
 -----------------------------------
 
Now we'll do the cube root of i:
we raise that to the {{{1/3}}}
power:
 
{{{(1(cos(pi((1+4n)/2))+i*sin(pi((1+4n)/2))))^(1/3)}}}
 
Now the rule for simplifying a complex number in trig
form is:
 
1. Raise the modulus (lenth of vector) to the power 
2. Multiply the argument (angle) by the power.
 
{{{1^(1/3)(cos((1/3)pi((1+4n)/2))+i*sin((1/3)pi((1+4n)/2)))}}}
 
which becomes:
 
{{{1(cos(pi((1+4n)/6))+i*sin(pi((1+4n)/6)))}}}
 
or
 
{{{cos(pi((1+4n)/6))+i*sin(pi((1+4n)/6))}}}
 
Now we let n take on any three consecutive integer values, but
the easiest are n=0, n=1 and n=2, so letting n=0:
 
{{{cos(pi((1+4*0)/6))+i*sin(pi((1+4*0)/6))}}}
 
{{{cos(pi/6)+i*sin(pi/6) = sqrt(3)/2+i*1/2}}}
 
That is this vector:
 
{{{drawing(300,300,-2,2,-2,2, line(sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.5,-.5,0,30) ),
graph(300,300,-2,2,-2,2),
line(sqrt(3)/2-.1,1/2,sqrt(3)/2,1/2), 
line(sqrt(3)/2,1/2-.1,sqrt(3)/2,1/2)
 

 )}}}
 
Now letting n=1:
 
{{{cos(pi((1+4*1)/6))+i*sin(pi((1+4*1)/6))}}}
 
{{{cos(pi(5/6))+i*sin(pi(5/6))}}}
 
{{{cos(5pi/6)+i*sin(5pi/6)=-sqrt(3)/2+(1/2)i}}}

That is this vector:
 
{{{drawing(300,300,-2,2,-2,2, line(-sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.5,-.5,0,150) ),
graph(300,300,-2,2,-2,2),
line(-sqrt(3)/2+.1,1/2-.1,-sqrt(3)/2,1/2), 
line(-sqrt(3)/2+.1,1/2+.05,-sqrt(3)/2,1/2)
 

 )}}}
 
Now letting n=2:
 
{{{cos(pi((1+4*2)/6))+i*sin(pi((1+4*2)/6))}}}
 
{{{cos(pi(9/6))+i*sin(pi(9/6))}}}

{{{cos(pi(3/2))+i*sin(pi(3/2))}}}

{{{cos(3pi/2)+i*sin(3pi/2)=0+(-1)i = -i}}}

That is this vector:
 
{{{drawing(300,300,-2,2,-2,2, line(-.1,-.7,0,-1),line(.1,-.7,0,-1), 
rectangle(0,0,0,-1), line(-.015,0,-.015,-1), blue(arc(0,0,.5,-.5,0,270) ),
graph(300,300,-2,2,-2,2), triangle(0,0,0,-1,0,0) )}}}

So the three cube roots of i are 
 
{{{sqrt(3)/2+(1/2)i}}}, {{{-sqrt(3)/2+(1/2)i}}}, and -i

Putting all three on the same graph:

{{{drawing(300,300,-2,2,-2,2, line(-sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.4,-.4,0,30) ),
graph(300,300,-2,2,-2,2),
line(-sqrt(3)/2+.1,1/2-.1,-sqrt(3)/2,1/2), 
line(-sqrt(3)/2+.1,1/2+.05,-sqrt(3)/2,1/2),
line(-.1,-.7,0,-1),line(.1,-.7,0,-1), 
rectangle(0,0,0,-1), line(-.015,0,-.015,-1), blue(arc(0,0,.5,-.5,0,150) ),
graph(300,300,-2,2,-2,2), triangle(0,0,0,-1,0,0),

line(sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.6,-.6,0,270) ),
graph(300,300,-2,2,-2,2),
line(sqrt(3)/2-.1,1/2,sqrt(3)/2,1/2), 
line(sqrt(3)/2,1/2-.1,sqrt(3)/2,1/2)
 )}}}


 
 
They form three equally spaced "spokes of a wheel":
 
{{{drawing(300,300,-2,2,-2,2, line(-sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.4,-.4,0,30) ),
graph(300,300,-2,2,-2,2),
line(-sqrt(3)/2+.1,1/2-.1,-sqrt(3)/2,1/2), 
line(-sqrt(3)/2+.1,1/2+.05,-sqrt(3)/2,1/2),
line(-.1,-.7,0,-1),line(.1,-.7,0,-1), 
rectangle(0,0,0,-1), line(-.015,0,-.015,-1), blue(arc(0,0,.5,-.5,0,150) ),
graph(300,300,-2,2,-2,2), triangle(0,0,0,-1,0,0),

line(sqrt(3)/2,1/2,0,0), 
blue(arc(0,0,.6,-.6,0,270) ),
graph(300,300,-2,2,-2,2),
line(sqrt(3)/2-.1,1/2,sqrt(3)/2,1/2), 
line(sqrt(3)/2,1/2-.1,sqrt(3)/2,1/2),circle(0,0,1)
 )}}}
 
Edwin</pre>