Question 207423
<pre><font size = 4 color = "indigo"><b>

I'll just do (2) and you can use it as a model to
do (1). 

{{{c = sqrt(2)-i*sqrt(2)}}}

Let's draw the picture of that complex number as a
right triangle on a graph. {{{c = x+iy=sqrt(2)-i*sqrt(2)}}}
where {{{x=sqrt(2)}}}, {{{y=-sqrt(2)}}}


{{{drawing(400,400,-2,2,-2,2,
graph(400,400,-2,2,-2,2), triangle(0,0,sqrt(2),-sqrt(2),sqrt(2),0),
locate(.7,.25,x=sqrt(2)), locate(1.45,-.6,y=-sqrt(2))

)}}}
</pre></font></b>
(a)  {{{abs(c)}}}.
<pre><font size = 4 color = "indigo"><b>
Sometimes this is called the "modulus". This is the length of 
the hypotenuse of that right triangle. We find it the same way 
we always find the hypotenuse, the Pythagorean theorem:

{{{abs(c) = r = sqrt((sqrt(2))^2+(-sqrt(2))^2)=sqrt(2+2)=sqrt(4)=2}}}

We label the hypotenuse {{{r=2}}} and this is the same as {{{abs(c)}}}

{{{drawing(400,400,-2,2,-2,2,
graph(400,400,-2,2,-2,2), triangle(0,0,sqrt(2),-sqrt(2),sqrt(2),0),
locate(.7,.25,x=sqrt(2)), locate(1.45,-.6,y=-sqrt(2)), 
locate(.4,-.7,r=2)

)}}}
</pre></font></b>
(b) find {{{arg(c)}}} when {{{-pi<arg(c)<pi}}}
<pre><font size = 4 color = "indigo"><b>
"Argument" means "angle" (you can remember it because "argument"
and "angle" both start with "a" and their third letters are "g")

Now since we are told that the argument {{{phi}}} is between {{{-pi}}}
and {{{""+pi}}}, we must take the argument as a negative angle 
measured by rotating clockwise from the right side of the x-axis, 
indicated by the blue arc below labeled {{{phi}}}, "phi".
It should be <font face = "symbol">q</font>, "theta", but I can't 
get that Greek letter on the notation program for this site, so 
I'll use {{{phi}}} instead.

{{{drawing(400,400,-2,2,-2,2, blue(arc(0,0,.8,-.8,315,360)),
graph(400,400,-2,2,-2,2), triangle(0,0,sqrt(2),-sqrt(2),sqrt(2),0),
locate(.7,.25,x=sqrt(2)), locate(1.45,-.6,y=-sqrt(2)), 
locate(.4,-.7,r=2), locate(.4,-.14,phi) 
)}}}

Since the triangle is an isosceles right triangle, its interior
angles are {{{45°}}} or {{{pi/4}}} in radians.  However since {{{-pi<arg(c)<pi}}}
tells us that the rotation is clockwise from the right
side of the x-axis, the angle is taken as negative. Thus we
take it as {{{phi=-pi/4}}} and we label it:

{{{drawing(400,400,-2,2,-2,2, blue(arc(0,0,.8,-.8,315,360)),
graph(400,400,-2,2,-2,2), triangle(0,0,sqrt(2),-sqrt(2),sqrt(2),0),
locate(.7,.25,x=sqrt(2)), locate(1.45,-.6,y=-sqrt(2)), 
locate(.4,-.7,r=2), locate(.4,-.06,phi=-pi/4) 
)}}}
</pre></font></b>
(c) write in polar form:
<pre><font size = 4 color = "indigo"><b>
Since {{{cos(phi)=x/r}}}, therefore {{{x=r*cos(phi)}}}

and since {{{sin(phi)=y/r}}}, therefore {{{y=r*sin(phi)}}}

So {{{x+iy=r*cos(phi)+i*r*sin(phi)}}}

and we can factor out {{{r}}} and get:

{{{x+iy=r(cos(phi)+i*sin(phi))}}}

This is the polar form {{{r(cos(phi)+i*sin(phi))}}} of {{{x+iy}}}

Therefore since {{{r=2}}} and {{{phi=-pi/4}}}

{{{c = sqrt(2)-i*sqrt(2)=2(cos(-pi/4)+i*sin(-pi/4))}}}
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(d) {{{c^6}}} or (sqrt(2)-i*sqrt(2))^6}}}
<pre><font size = 4 color = "indigo"><b>
DeMoivre's theorem says:

{{{(x+iy)^n = (r(cos(phi)+i*sin(phi)))^n = r^n*(cos(n*phi)+i*sin(n*phi))}}}

Therefore:

{{{c^6=(sqrt(2)-i*sqrt(2))^6 = (2(cos(-pi/4)+i*sin(-pi/4)))^6 = 2^6*(cos(6*(-pi/4))+i*sin(6*(-phi/4)))=64(cos(-3pi/2)+i*sin(-3pi/2))}}}

Since {{{-3pi/2}}} is coterminal with {{{pi/2}}}, we have

{{{c^6=64(cos(pi/2)+i*sin(pi/2))=64(0+i*1)= 64(i) = 64i}}}

Edwin</pre>