Roots of Complex Numbers
In[30]:=
z=3+4 I
Sqrt[z]
Out[30]=
3 + 4 I
Out[31]=
2 + I
In[32]:=
Sqrt[Abs[z]] Exp[I Arg[z]/2]
N[Sqrt[Abs[z]] Exp[I Arg[z]/2]]
Out[32]=
I/2 ArcTan[3, 4]
Sqrt[5] E
Out[33]=
2. + 1. I
Shouldn't there be two square roots? If s is a square root, isn't -s also?
In[34]:=
z=3+4 I
s=-Sqrt[z]
Out[34]=
3 + 4 I
Out[35]=
-2 - I
In[36]:=
s^2
Out[36]=
3 + 4 I
Look at the plot of the two roots. Do you notice anything?
In[37]:=
p1=ListPlot[{{2,1},{-2,-1}}, PlotStyle->{RGBColor[1,0,0],PointSize[.02]}]

Out[38]=
-Graphics-
Look again.
In[39]:=
p2=PolarPlot[Sqrt[Abs[z]],{t, 0, 2 Pi},DisplayFunction->Identity];
Show[p1,p2, DisplayFunction->$DisplayFunction,AspectRatio->Automatic]

Out[40]=
-Graphics-
The two square roots are Pi away from each other on the circle of radius Sqrt[ |z| ].
Let's look at cube roots
In[41]:=
z
c1=N[z^(1/3)]
Out[41]=
3 + 4 I
Out[42]=
1.62894 + 0.520175 I
There should be three cube roots. The other two are (more on this later):
In[43]:=
c2=N[Abs[z]^(1/3) Exp[I(Arg[z]/3+(2 Pi/3))]]
c3=N[Abs[z]^(1/3) Exp[I(Arg[z]/3+(4 Pi/3))]]
Out[43]=
-1.26495 + 1.15061 I
Out[44]=
-0.363984 - 1.67079 I
Don't believe me?
In[45]:=
c2^3
c3^3
Out[45]=
3. + 4. I
Out[46]=
3. + 4. I
In[47]:=
p3=ListPlot[{{1.62894,.520175},{-1.26495,1.15061},
{-.363984,-1.67079}},DisplayFunction->Identity,
PlotStyle->{RGBColor[1,0,0],PointSize[.02]}];
p4=PolarPlot[Abs[z]^(1/3),{t,0, 2 Pi},DisplayFunction->Identity];
Show[p3,p4,DisplayFunction->$DisplayFunction,
AspectRatio->Automatic]

Out[48]=
-Graphics-
The cube roots are evenly spaced around the circle of radius |z|^(1/3).
In general, the nth roots are evenly spaced around the circle of radius |z|^(1/n), starting with the "first", or principal root. Putting this together with the polar form, one can also see that the n nth roots of z are given by
|z|^(1/n) Exp[i (theta/n + 2kPi/n)], k=0, 1, 2, 3, . . ., n-1
Exercise: Verify using Euler's equation that the n numbers given above are indeed nth roots of z.
Up to Complex Numbers