Roots of Complex Numbers (Sharp, TI, Casio)

12302022, 10:58 PM
Post: #1




Roots of Complex Numbers (Sharp, TI, Casio)
Hi all.
On a whim (and maybe a hunch, but a damn lucky guess), I keyed in a method that calculates nth roots of Complex Numbers on the TI36X Pro/30X Pro MathPrint Sharp ELW516X 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 TI30X 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 nth 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? 

12312022, 12:27 PM
Post: #2




RE: Roots of Complex Numbers (Sharp, TI, Casio)
z = z * e^(θ*i), where θ = arg(z)
(a^b)^c = a^(b*c) → z^k = z^k * e^(kθ*i) (12302022 10:58 PM)Matt Agajanian Wrote: Example Calculate 4th root of (15625+0.719413999i) 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) 

12312022, 07:26 PM
(This post was last modified: 12312022 07:28 PM by toml_12953.)
Post: #3




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(12302022 10:58 PM)Matt Agajanian Wrote: Example Calculate 4th root of (15625+0.719413999i) On Prime I get this: Tom L Cui bono? 

12312022, 08:02 PM
Post: #4




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(12312022 07:26 PM)toml_12953 Wrote: On Prime I get this: WolframAlpha returns the same results. https://www.wolframalpha.com/input?i2d=t...9%2C.25%5D 

12312022, 08:26 PM
Post: #5




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(12312022 12:27 PM)Albert Chan Wrote: z = z * e^(θ*i), where θ = arg(z) Thanks! Now that I know, how should I have entered my original expression in the 30X Pro MathPrint and 36X Pro? Thanks 

12312022, 11:41 PM
(This post was last modified: 01012023 01:43 AM by klesl.)
Post: #6




RE: Roots of Complex Numbers (Sharp, TI, Casio)
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 TI30X 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 nth 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 nth root with 2 steps only https://www.youtube.com/watch?v=7gWJEZgohAk 

12312022, 11:58 PM
(This post was last modified: 01012023 12:11 AM by Matt Agajanian.)
Post: #7




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(12312022 11:41 PM)klesl Wrote: step by step solutions by using DeMoivre's theorem 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? 

01012023, 01:25 AM
Post: #8




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(11+2i)^4=11753+10296*i
it is pretty clear 11+2i is not result of your example (15625+0.71941399i)^(1/4) 

01012023, 05:34 AM
Post: #9




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(12312022 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 { 28Byte Prgm } (12302022 10:58 PM)Matt Agajanian Wrote: Example Calculate 4th root of (15625+0.719413999i) 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 \(r \cdot e^{it}\). 

01012023, 08:48 AM
Post: #10




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01012023 05:34 AM)Thomas Klemm Wrote:(12312022 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? 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. 

01012023, 10:18 AM
(This post was last modified: 01012023 10:22 AM by Thomas Klemm.)
Post: #11




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01012023 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. 

01012023, 10:03 PM
(This post was last modified: 01022023 05:58 AM by Matt Agajanian.)
Post: #12




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01012023 10:18 AM)Thomas Klemm Wrote:(01012023 08:48 AM)Matt Agajanian Wrote: I'd just like to know how it can be accomplished on the TI models. 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? 

01022023, 08:47 AM
Post: #13




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01012023 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:


01022023, 10:35 PM
Post: #14




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01022023 08:47 AM)Thomas Klemm Wrote:(01012023 10:03 PM)Matt Agajanian Wrote: Correct? 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. 

01032023, 08:44 PM
(This post was last modified: 01042023 03:21 AM by Matt Agajanian.)
Post: #15




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01022023 08:47 AM)Thomas Klemm Wrote:(01012023 10:03 PM)Matt Agajanian Wrote: Correct? Sorry for the doublepost. 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) ? 

01052023, 12:29 AM
(This post was last modified: 01052023 02:40 AM by Matt Agajanian.)
Post: #16




RE: Roots of Complex Numbers (Sharp, TI, Casio)
(01022023 08:47 AM)Thomas Klemm Wrote:(01012023 10:03 PM)Matt Agajanian Wrote: Correct? 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 selfexplanatory. 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 PRy(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 

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