Question 130377
 Use De Moivre's Formula to find (-1+2i)^36
<pre><b><font size = 4>
To find (x+yi)<sup>n</sup>
First change x+yi to trigonometric form.

To change x + iy to trigonometric form, we plot
the point (x,y), and draw a right triangle with
vertices (x,y), (x,0), and (0,0)
We calculate the hypotenuse r from r² = x² + y²,
which is called the modulus or absolute value
of the complex number x+yi
We calculate the angle <font face = "symbol">q</font>
from either

tan<font face = "symbol">q</font> = {{{y/x}}},
or 
sin<font face = "symbol">q</font> = {{{y/r}}},
or
cos<font face = "symbol">q</font> = {{{x/r}}},

Then x + yi has the trigonometric form:

x + yi = r(cos<font face = "symbol">q</font> + i·sin<font face = "symbol">q</font>)

Then we use the formula:

[r(cos<font face = "symbol">q</font> + i·sin<font face = "symbol">q</font>)]<sup>n</sup> = r<sup>n</sup>[cos(n<font face = "symbol">q</font>) + i·sin(n<font face = "symbol">q</font>)]

The real part of -1+2i is -1
The imaginary part of -1+2i is 2, the coefficient of i.

We plot the point (-1,2), and draw a right triangle with
vertices (-1,2), (-1,0), and (0,0)

{{{drawing(500,500, -2.4,2.4,-2.4,2.4,
locate(-1.3,1,"y=2"), locate(-.7,.2,"x=-1"),
graph(500,500,-2.4,2.4,-2.4,2.4),
triangle(0,0,-1,0,-1,2),
locate(-1.3,2.2,"(-1,2)")
 )}}}

Next we calculate r:

r² = x² + y²
r² = (-1)² + 2²
r² = 1 + 4
r² = 5
r = {{{sqrt(5)}}}

{{{drawing(500,500, -2.4,2.4,-2.4,2.4,
locate(-1.3,1,"y=2"), locate(-.7,.2,"x=-1"),
graph(500,500,-2.4,2.4,-2.4,2.4),
locate(-.5,1.3,"r="), locate(-.35,1.3,sqrt(5)),
triangle(0,0,-1,0,-1,2),
locate(-1.3,2.2,"(-1,2)")
 )}}}

Now we need to calculate the angle <font face = "symbol">q</font>
indicated by the arc:

{{{drawing(500,500, -2.4,2.4,-2.4,2.4,
locate(-1.3,1,"y=2"), locate(-.7,.2,"x=-1"),
graph(500,500,-2.4,2.4,-2.4,2.4,sqrt(.6-x^2)sqrt(.3+x)/sqrt(.3+x)),
locate(-.5,1.3,"r="), locate(-.35,1.3,sqrt(5)),
triangle(0,0,-1,0,-1,2),
locate(-1.3,2.2,"(-1,2)")
 )}}}

We first calculate its reference angle from, say

tan(ref<font face = "symbol">q</font>) = {{{2/1}}},

So we find the reference angle from the inverse tangent
on the calculator:

ref<font face = "symbol">q</font> = 63.4349°

Now we know that <font face = "symbol">q</font> is in 
Quadrant II, so we subtract the reference angle from 
180° and get <font face = "symbol">q</font> = 116.5651°

So the trigonometric form is
           
-1 + 2i = {{{sqrt(5)}}}[cos(116.5651°)+ i·sin(116.5651°)]

So using the formula

[r(cos<font face = "symbol">q</font> + i·sin<font face = "symbol">q</font>)]<sup>n</sup> = r<sup>n</sup>[cos(n<font face = "symbol">q</font>) + i·sin(n<font face = "symbol">q</font>)]

[{{{sqrt(5)}}}(cos116.5651° + i·sin116.5651°)<sup>36</sup> =
{{{(sqrt(5))^36}}}[cos(36·116.5651°) + i·sin(36·116.5651°)] = 
{{{(5^(1/2))^36}}}[cos(4196.3436°) + i·sin(4196.3436°)] = 
{{{5^18}}}[cos(4196.3436°) + i·sin(4196.3436°)]

Now we can replace the angle 4196.3436° by its smallest 
positive coterminal angle.

To do this we divide the angle by 360° and get 11.65651.  The
whole part is 11 so we subtract 360° 11 times, which means we 
subtract 11·360° or 3960° from 4196.3436° and get

4196.3436° - 3960° = 236.3436°

So the answer in trig form is

5<sup>18</sup>(cos236.3436° + i·sin236.3436°)

Using a calculator we get

-2.114 - 3.175i for the answer in standard form.

Edwin</pre>