Roots of Complex Numbers (Sharp, TI, Casio)

[Matt Agajanian]

Hi all.

On a whim (and maybe a hunch, but a damn lucky guess), I keyed in a method that calculates n-th roots of Complex Numbers on the
TI-36X Pro/30X Pro MathPrint
Sharp EL-W516X and T
Casio 115ES and 991 line.

It goes like this:

Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

TI-30X Pro MathPrint:
[math] [P→Rx] (x^.25,y/4) → 11 [Real part]
[math] [P→Ry] (x^.25,y/4) → 2 [Imaginary part]
Thus, the answer is 11+2i

General algorithm:
Calculate n-th root of x+yi

[math] [P→Rx] (x^(1/n),y/n) → a [Real part]
[math] [P→Ry] (x^(1/n),y/n) → b [Imaginary part]

Casio & Sharp models would then follow the above algorithm.

But then I thought maybe the algorithm for Raising i to integer powers could be modified as an alternative method.

For a reminder,

Example 1: (11 + 2i)^4 = 11753 + 10296i

(Radian Mode)
R>Pr(11,2)^4 sto→ x (15625)
R>PΘ(11,2)*4 sto→ y (0.719413999)
P>Rx(x,y) returns 11753
P>Ry(x,y) returns 10296

So, wouldn't just switching to this:

Example: Calculate the 4th root of a+bi
(Radian Mode)
R>Pr(a,b)^(1/4) sto x
R>PΘ(a,b)/4 sto y
P>Rx(x,y)
P>Ry(x,y)

or would this work:

(Radian Mode)
R>Pr(a^(1/4),b/4) sto x
R>PΘ(a^(1/4),b/4) sto y
P>Rx(x,y)
P>Ry(x,y)
?

or the first algorithm noted above the accurate and only method?

[Albert Chan]

z = |z| * e^(θ*i), where θ = arg(z)

(a^b)^c = a^(b*c)      → z^k = |z|^k * e^(kθ*i)

(12-30-2022 10:58 PM)Matt Agajanian Wrote:  Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

TI-30X Pro MathPrint:
[math] [P→Rx] (x^.25,y/4) → 11 [Real part]
[math] [P→Ry] (x^.25,y/4) → 2 [Imaginary part]
Thus, the answer is 11+2i

This example is in error, since (11 + 2i)^4 = 11753 + 10296i
This is what it mean:

(15625 * e^(0.719413999i)) ^ 0.25
= (15625 ^ 0.25) * e^(0.719413999i * 0.25)
≈ (11 + 2i)

[toml_12953]

(12-30-2022 10:58 PM)Matt Agajanian Wrote:  Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

Thus, the answer is 11+2i

On Prime I get this:

[Steve Simpkin]

(12-31-2022 07:26 PM)toml_12953 Wrote:  On Prime I get this:
[ image ]

WolframAlpha returns the same results.
   

https://www.wolframalpha.com/input?i2d=t...9%2C.25%5D

[Matt Agajanian]

(12-31-2022 12:27 PM)Albert Chan Wrote:  z = |z| * e^(θ*i), where θ = arg(z)

(a^b)^c = a^(b*c)      → z^k = |z|^k * e^(kθ*i)

(12-30-2022 10:58 PM)Matt Agajanian Wrote:  Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

TI-30X Pro MathPrint:
[math] [P→Rx] (x^.25,y/4) → 11 [Real part]
[math] [P→Ry] (x^.25,y/4) → 2 [Imaginary part]
Thus, the answer is 11+2i

This example is in error, since (11 + 2i)^4 = 11753 + 10296i
This is what it mean:

(15625 * e^(0.719413999i)) ^ 0.25
= (15625 ^ 0.25) * e^(0.719413999i * 0.25)
≈ (11 + 2i)

Thanks!

Now that I know, how should I have entered my original expression in the 30X Pro MathPrint and 36X Pro?

Thanks

[klesl]

step by step solutions by using DeMoivre's theorem
https://www.emathhelp.net/calculators/al...13999i&n=4
or video
https://www.youtube.com/watch?v=o6bUy4Vg7yM
So similarly for TI-30X Pro:
1. step: enter complex number 15625+0.719413999i
2. get magnitude and angle, optionally you can store these values to memory, e.g. magnitude to x and angle to y
3. enter n-th root from magnitude, enter angle symbol (menu "complex" - option 1), enter angle divided by the n
4 press enter to get result
for n=4:
x^(1/4)<y/4
11.18033989+0.000128693i

n-th root with 2 steps only
https://www.youtube.com/watch?v=7gWJEZgohAk

[Matt Agajanian]

(12-31-2022 11:41 PM)klesl Wrote:  step by step solutions by using DeMoivre's theorem
https://www.emathhelp.net/calculators/al...13999i&n=4

Yes. Thanks. But, is a way to arrive at the final answers by just using the 36X Pro and 30X Pro MathPrint without all the gymnastics?

But what function am I using when I put (15625+0.719413999i)^(1/4) into the 42S and get 11+2i as a single result?

[klesl]

(11+2i)^4=11753+10296*i
it is pretty clear 11+2i is not result of your example (15625+0.71941399i)^(1/4)

[Thomas Klemm]

(12-31-2022 11:58 PM)Matt Agajanian Wrote:  But what function am I using when I put (15625+0.719413999i)^(1/4) into the 42S and get 11+2i as a single result?

Code:

00 { 28-Byte Prgm }
01 RAD
02 POLAR
03 15625
04 0.71941399
05 COMPLEX
06 4
07 1/X
08 Y↑X
09 RECT
10 END

---

(12-30-2022 10:58 PM)Matt Agajanian Wrote:  Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

(…)

Example 1: (11 + 2i)^4 = 11753 + 10296i

(Radian Mode)
R>Pr(11,2)^4 sto→ x (15625)
R>PΘ(11,2)*4 sto→ y (0.719413999)

I assume that part of the confusion stems from the fact that you use variables \(x\) and \(y\) for radius and angle.
Thus I suggest to use \(r\) and \(t\) instead.
Hopefully this prevents you from writing \(r + it\) when you mean \(r \cdot e^{it}\).

[Matt Agajanian]

(01-01-2023 05:34 AM)Thomas Klemm Wrote:  

(12-31-2022 11:58 PM)Matt Agajanian Wrote:  But what function am I using when I put (15625+0.719413999i)^(1/4) into the 42S and get 11+2i as a single result?

Code:

00 { 28-Byte Prgm }
01 RAD
02 POLAR
03 15625
04 0.71941399
05 COMPLEX
06 4
07 1/X
08 Y↑X
09 RECT
10 END

---

(12-30-2022 10:58 PM)Matt Agajanian Wrote:  Example Calculate 4th root of (15625+0.719413999i)

Let x= 15625, y=0.719413999

(…)

Example 1: (11 + 2i)^4 = 11753 + 10296i

(Radian Mode)
R>Pr(11,2)^4 sto→ x (15625)
R>PΘ(11,2)*4 sto→ y (0.719413999)

I assume that part of the confusion stems from the fact that you use variables \(x\) and \(y\) for radius and angle.
Thus I suggest to use \(r\) and \(t\) instead.
Hopefully this prevents you from writing \(r + it\) when you mean \(r \cdot e^{it}\).

Yes. I can see and understand what to do on the 42S since it can handle the whole range of transcendental functions in complex mode. And I have entered the (11+2i)^4 as well as calculating the fourth root of the previous result. But, switching it around to accomplishing the same functions and results on the 36X Pro and 30X Pro MathPrint, how would I do that?

Yes, I can use the 42 instead of the 30 and 36. I'd just like to know how it can be accomplished on the TI models.

[Thomas Klemm]

(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.

[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?

[Thomas Klemm]

(01-01-2023 10:03 PM)Matt Agajanian Wrote:  Correct?

This looks good to me.
But now both blocks became identical.

You could use variables \(x\) and \(y\) instead of \(a\) and \(b\) if that helps for readability.
And another variable \(n\) for the exponent instead of \(4\):

R>Pr(x, y)^n sto r
R>PΘ(x, y)*n sto t
P>Rx(r, t)
P>Ry(r, t)

Now you can use it for any of these cases:

• \(n = 4\)
• \(n = -4\)
• \(n = \frac{1}{4}\)
  

[Matt Agajanian]

(01-02-2023 08:47 AM)Thomas Klemm Wrote:  

(01-01-2023 10:03 PM)Matt Agajanian Wrote:  Correct?

This looks good to me.
But now both blocks became identical.

You could use variables \(x\) and \(y\) instead of \(a\) and \(b\) if that helps for readability.
And another variable \(n\) for the exponent instead of \(4\):

R>Pr(x, y)^n sto r
R>PΘ(x, y)*n sto t
P>Rx(r, t)
P>Ry(r, t)

Now you can use it for any of these cases:

• \(n = 4\)
• \(n = -4\)
• \(n = \frac{1}{4}\)
  

My goof.

I mean to say

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

for the second example. Anyway, thanks for the catch all process. Yeah, even though my samples were specific values, it was just making an example.

[Matt Agajanian]

(01-02-2023 08:47 AM)Thomas Klemm Wrote:  

(01-01-2023 10:03 PM)Matt Agajanian Wrote:  Correct?

This looks good to me.
But now both blocks became identical.

You could use variables \(x\) and \(y\) instead of \(a\) and \(b\) if that helps for readability.
And another variable \(n\) for the exponent instead of \(4\):

R->Pr(x, y)^n sto r
R->PΘ(x, y)*n sto t
P->Rx(r, t)
P->Ry(r, t)

Now you can use it for any of these cases:

• \(n = 4\)
• \(n = -4\)
• \(n = \frac{1}{4}\)
  

Sorry for the double-post. Another thought.

When I use this method for P->R, is it


P->Rx(r, Θ)^n sto r
P->Ry(x, Θ)*n sto t
R->Pr(r, t)
R->Pθ(r, t)

or

r^n STO a
θ*n STO b
P->Rx(a,b)
P->Ry(a,b)


?

[Matt Agajanian]

(01-02-2023 08:47 AM)Thomas Klemm Wrote:  

(01-01-2023 10:03 PM)Matt Agajanian Wrote:  Correct?

This looks good to me.
But now both blocks became identical.

You could use variables \(x\) and \(y\) instead of \(a\) and \(b\) if that helps for readability.
And another variable \(n\) for the exponent instead of \(4\):

R>Pr(x, y)^n sto r
R>PΘ(x, y)*n sto t
P>Rx(r, t)
P>Ry(r, t)

Now you can use it for any of these cases:

• \(n = 4\)
• \(n = -4\)
• \(n = \frac{1}{4}\)
  

Excellent techniques

I’ve been working on some new routines.

Please let me know if any of these are legit:

Even if you're not familiar with the TI 36X Pro or 30X MathPrint, the keystrokes should be self-explanatory.

Complex Number Power & Root Calculations
(x+yi)^n
Example: (11+2i)^4
R→Pr(11,2)^4 (15625)
sto x
R→Pθ(11,2)*4 (0.719413999)
sto y
P→Rx(x,y) 11753
P→Ry(x,y) 10296

Fourth root of r∡θ [(r∡θ)^1/4]
Example: (15625∡ 0.719413999)^1/4
15625^1/4 11.18033989
sto x
0.719413999/4 0.17985335
sto y
P→Rx(x,y) 11
P-Ry(x,y) 2

Fourth root of (x+yi) [(x+yi)^1/4]
Example: (11753+10296 i)^1/4
11753
sto x
10296
sto y
R→Pr(x,y)^.25
sto a
R→Pθ(x,y)/4
sto b
P→Rx(a,b) 11
P→Ry(a,b) 2

(r∡θ)^n
Example: (11.18033989∡0.1798535)^4 (11+2i)^4 in polar form)
(11.18033989∡0.1798535)^4
11.18033989
sto x
0.1798535
sto y
P→Rx(x^4,y*4) → 11753
sto a
P→Ry(x^4,y*4) → 10296
sto b
R→Pr(a,b) → 15625
R→Pθ(a,b) → 0.71914

Thanks