Roots of Complex Numbers (Sharp, TI, Casio)

[Matt Agajanian]

(01-01-2023 10:18 AM)Thomas Klemm Wrote:  

(01-01-2023 08:48 AM)Matt Agajanian Wrote:  I'd just like to know how it can be accomplished on the TI models.

I think that you are already close.

Let's assume:

\(
\begin{align}
z
&= 11753 + 10296i \\
&= a + ib \\
\\
&= 15625 \, \measuredangle \, 0.719413999 \\
&= u \, \measuredangle \, v \\
\end{align}
\)

Rectangular Coordinates

Here we use \(a + ib = 11753 + 10296i\):

(Radian Mode)
R>Pr(a, b)^(1/4) sto r
R>PΘ(a, b)/4 sto t
P>Rx(r, t)
P>Ry(r, t)

Polar Coordinates

Here we use \(u \, \measuredangle \, v = 15625 \, \measuredangle \, 0.719413999\):

(Radian Mode)
u^(1/4) sto r
v/4 sto t
P>Rx(r, t)
P>Ry(r, t)

---

I must admit that I have no idea how this calculator works.
Thus I could be totally wrong.

With this method, I would presume

(Radian Mode)
R>Pr(a, b)^4 sto r
R>PΘ(a, b)*4 sto t
P>Rx(r, t)
P>Ry(r, t)

and

R>Pr(a, b)^4 sto r
R>PΘ(a, b)*4 sto t
P>Rx(r, t)
P>Ry(r, t)

Would do the reverse and raise a+bi to the fourth power. Correct?