Complex cube root

Introduction

Complex number W = complex number w³. Origin at point $(0,0)$. $w_{real},W_{real}$ parallel to $X$ axis. $w_{imag},W_{imag}$ parallel to $Y$ axis. $w_{real}=1.2;\ w_{imag}=0.5;\ w_{mod}={\sqrt {1.2^{2}+0.5^{2}}}=1.3.$ $W_{real}=0.828;\ ;\ W_{imag}=2.035;$ $W_{mod}={\sqrt {0.828^{2}+2.035^{2}}}=2.197.$ $W_{mod}=w_{mod}^{3}=1.3^{3}=2.197.$ $W_{\phi }=3w_{\phi }.$ By cosine triple angle formula: $\cos W_{phi}=4\cdot ({\frac {1.2}{1.3}})^{3}-3\cdot {\frac {1.2}{1.3}}={\frac {828}{2197}}={\frac {W_{real}}{W_{mod}}}.$ See "3 cube roots of W" in Gallery below. Let complex numbers $W=$ a $+$ b$\cdot i$ and $w=$ p $+$ q$\cdot i.$ Let $W=w^{3}.$ When given $a,b,$ aim of this page is to calculate $p,q.$ In the diagram complex number $w=p+qi=w_{real}+i\cdot w_{imag}=w_{mod}(\cos w_{\phi }+i\cdot \sin w_{\phi }),$ where $w_{real},w_{imag}$ are the real and imaginary components of $w,$ the rectangular components. $w_{mod},w_{\phi }$ are the modulus and phase of $w,$ the polar components. Similarly, $W_{real},W_{imag},W_{mod},W_{\phi }$ are the corresponding components of $W.$ $W=w^{3}=(w_{mod}(\cos w_{\phi }+i\cdot \sin w_{\phi }))^{3}$ $=w_{mod}^{3}(\cos(3w_{\phi })+i\cdot \sin(3w_{\phi }))$ $=W_{mod}(\cos(W_{\phi })+i\cdot \sin(W_{\phi }))$ $=W$ There are 2 significant calculations: $w_{mod}={\sqrt[{3}]{W_{mod}}}$ and $\cos w_{\phi }=\cos {\frac {W_{\phi }}{3}}.$

Implementation

Cos φ from cos 3φ Graph of $\cos(3A)=4\cos ^{3}(A)-3\cos(A)$ Formula and/or graph are used to calculate $\cos(A)$ if $\cos(3A)$ is given. The cosine triple angle formula is: $\cos(3\theta )=4\cos ^{3}\theta -3\cos \theta .$ This formula, of form $y=4x^{3}-3x,$ permits $\cos(3\theta )$ to be calculated if $\cos \theta$ is known. If $\cos(3\theta )$ is known and the value of $\cos \theta$ is desired, this identity becomes: $4\cos ^{3}\theta -3\cos \theta -\cos(3\theta )=0.$ $\cos \theta$ is the solution of this cubic equation. In fact this equation has three solutions, the other two being $\cos(\theta \pm 120^{\circ }).$ $\cos(3(\theta \pm 120^{\circ }))=\cos(3\theta \pm 360^{\circ })=\cos(3\theta ).$ It is sufficient to calculate only $\cos \theta$ from $\cos 3\theta ,$ accomplished by the following code: # python code cosAfrom_cos3Adebug = 0 def cosAfrom_cos3A(cos3A) : cos3A = Decimal(str(cos3A))+0 if 1 >= cos3A >= -1 : pass else : print ('cosAfrom_cos3A(cos3A) : cos3A not in valid range.') return None ''' if cos3A == 0 : A = 90 and cos3A = cosA if cos3A == 1 : A = 0 and cos3A = cosA if cos3A == -1 : A = 180 and cos3A = cosA ''' if cos3A in (0,1,-1) : return cos3A # From the cosine triple angle formula: a,b,c,d = 4,0,-3,-cos3A # prepare for Newton's method. if d < 0 : x = Decimal(1) else : x = -Decimal(1) count = 31; L1 = [x]; almostZero = Decimal('1e-' + str(prec-2)) # Newton's method: while 1 : count -= 1 if count <= 0 : print ('cosAfrom_cos3A(cos3A): count expired.') break y = a*x*x*x + c*x + d if cosAfrom_cos3Adebug : print ('cosAfrom_cos3A(cos3A) : x,y =',x,y) if abs(y) < almostZero : break slope = 12*x*x + c delta_x = y/slope x -= delta_x if x in L1[-1:-5:-1] : # This value of x has been used previously. print ('cosAfrom_cos3A(cos3A): endless loop detected.') break L1 += [x] return x When $x==1,$ slope $=12x^{2}-3=9.$ Within area of interest, maximum absolute value of slope $=9,$ a rather small value for slope. Consequently, with only 9 passes through loop, Newton's method produces a result accurate to 200 places of decimals . There are 3 conditions, any 1 of which terminates the loop: abs(y) very close to 0 (normal termination). count expired. endless loop detected with the same value of x repeated. Calculation of cube roots of W Calculation of 1 root # python code. def complexCubeRoot (a,b) : ''' p,q = complexCubeRoot (a,b) where: * a,b are rectangular coordnates of complex number a + bi. * (p + qi)^3 = a + bi. ''' a = Decimal(str(a)) b = Decimal(str(b)) if a == b == 0 : return (Decimal(0), Decimal(0)) if a == 0 : return (Decimal(0), -simpleCubeRoot(b)) if b == 0 : return (simpleCubeRoot(a), Decimal(0)) Wmod = (a*a + b*b).sqrt() wmod = simpleCubeRoot (Wmod) cosWφ = a/Wmod coswφ = cosAfrom_cos3A(cosWφ) p = coswφ * wmod # wreal q = (wmod*wmod - p*p).sqrt() # wimag # Resolve ambiguity of q, + or -. v1 = - 3*p*p*q + q*q*q qp = abs(b + v1 ) qn = abs(b - v1) if qp > qn : q *= -1 return p,q For function simpleCubeRoot() see Cube_root#Implementation Calculation of 3 roots See Cube roots of unity. The cube roots of unity are : $1,{\frac {-1\pm i{\sqrt {3}}}{2}}.$ When $r_{0}={\sqrt[{3}]{W}}$ is known, the other 2 cube roots are: $r_{1}=r_{0}\cdot {\frac {-1+i{\sqrt {3}}}{2}}$ $r_{2}=r_{0}\cdot {\frac {-1-i{\sqrt {3}}}{2}}$ # python code def complexCubeRoots (a,b) : ''' r0,r1,r2 = complexCubeRoots (a,b) where: * a,b are rectangular coordnates of complex number a + bi. * (p + qi)^3 = a + bi. * r0 = (p0,q0) * r1 = (p1,q1) * r2 = (p2,q2) ''' p,q = complexCubeRoot (a,b) r3 = Decimal(3).sqrt() # Square root of 3. pr3,qr3 = p*r3, q*r3 # r1 = ((-p-q*r)/2, (p*r - q)/2) # r2 = ((-p+q*r)/2, (-p*r - q)/2) r0 = (p,q) r1 = ((-p-qr3)/2, (pr3 - q)/2) r2 = ((-p+qr3)/2, (-pr3 - q)/2) return r0,r1,r2

Testing results

$(p+q\cdot i)^{3}=p^{3}-3pq^{2}+(3p^{2}q-q^{3})\cdot i=a+b\cdot i.$ $a=p^{3}-3pq^{2};\ b=3p^{2}q-q^{3}.$ # Python code used to test results of code above, def complexCubeRootsTest (a,b) : print ('\n++++++++++++++++++++') print ('a,b =', a,b) almostZero = Decimal('1e-' + str(prec-5)) r0,r1,r2 = complexCubeRoots (a,b) for root in (r0,r1,r2) : p,q = root print (' pq =',(p), (q)) a_,b_ = (p*p*p - 3*p*q*q), (3*p*p*q - q*q*q) if a : v1 = abs((a_-a)/a) if v1 > almostZero : print ('error *',a_,a,v1) else : v1 = abs(a_) if v1 > almostZero : print ('error !',a_,a,v1) if b : v1 = abs((b_-b)/b) if v1 > almostZero : print ('error &',b_,b,v1) else : v1 = abs(b_) if v1 > almostZero : print ('error %',b_,b,v1) return r0,r1,r2 import decimal Decimal = D = decimal.Decimal prec = decimal.getcontext().prec # precision cosAfrom_cos3Adebug = 1 for p in range (-10,11,1) : for q in range (-10,11,1) : a = p*p*p - 3*p*q*q b = 3*p*p*q - q*q*q complexCubeRootsTest(a,b)

Gallery

Cube roots of 8i. Cube roots of -8. 3 cube roots of W.

Method #2 (Graphical)

Introduction

Graphs of 2 curves showing complex cube roots as points of intersection of the 2 curves.

Let complex number $w=p+qi.$

Then $W=w^{3}=(p+qi)^{3}=p^{3}-3pq^{2}+(3p^{2}q-q^{3})i.$

Let $W=a+bi.$

Then:

$a=p^{3}-3pq^{2}$ and

$b=3p^{2}q-q^{3}.$

When $W$ is given and $w$ is desired, $w$ may be calculated from the solutions of 2 simultaneous equations:

$p^{3}-3pq^{2}-a=0\ \dots \ (1)$ and

$3p^{2}q-q^{3}-b=0\ \dots \ (2).$

For example, let $W=(39582+3799i).$

Then equations $(1)$ and $(2)$ become (for graphical purposes):

$x^{3}-3xy^{2}-39582=0\ \dots \ (3),$ black curve in diagram, and

$3x^{2}y-y^{3}-3799=0\ \dots \ (4),$ red curve in diagram.

Three points of intersection of red and black curves are:

$(-18,29),$

$(34.11473670974872,1.0884572681198943),$ and

$(-16.11473670974872,-30.088457268119896),$

interpreted as the three complex cube roots of $W,$ namely:

$w_{0}=(-18+29i),$

$w_{1}=(34.11473670974872+1.0884572681198943i)$ and

$w_{2}=(-16.11473670974872-30.088457268119896i).$

Proof: # python code # Each cube root cubed. >>> w0 = (-18 + 29j) >>> w1 = (34.11473670974872 + 1.0884572681198943j) >>> w2 = (-16.11473670974872 - 30.088457268119896j) >>> for w in (w0,w1,w2) : w**3 ... (39582+3799j) (39582+3799j) (39582+3799j) # The moduli of all 3 cube roots. >>> for w in (w0,w1,w2) : (w.real**2 + w.imag**2) ** 0.5 ... 34.132096331752024 34.132096331752024 34.132096331752024

Preparation

This method depends upon selection of most appropriate quadrant.

In the example above, quadrant $2$ is chosen because any non-zero positive value of $y$ intersects red curve and any non-zero negative value of $x$ intersects black curve.

Figures 1-4 below show all possibilities of $\pm a$ and $\pm b.$

Figure 1. When both $a,b$ are positive, quadrant 2 is chosen. $(-18+29i)^{3}$$=(39582+3799i)$ Figure 2. When $a$ is positive and $b$ is negative, quadrant 3 is chosen. $(-18-29i)^{3}$$=(39582-3799i)$ Figure 3. When $a$ is negative and $b$ is positive, quadrant 1 is chosen. $(18+29i)^{3}$$=(-39582+3799i)$ Figure 4. When both $a,b$ are negative, quadrant 4 is chosen. $(18-29i)^{3}$$=(-39582-3799i)$

Description of method

Four points

Graphs of 2 curves showing 4 points that enclose one of the complex cube roots. Area enclosed by the 4 points becomes progressively smaller and smaller until the point of intersection is identified. Assume that $W=-39582-3799i,$ in which case both $a,b$ are negative and quadrant $4$ is chosen. In quadrant $4$ any non-zero positive value of x intersects black curve and any non-zero negative value of y intersects red curve. Choose any convenient negative, non-zero value of $y.$ Let $y=-18.$ Using this value of $y,$ calculate coordinates of point $A$ on red curve. Using $x$ coordinate of point $A,$ calculate coordinates of point $B$ on black curve. Using $y$ coordinate of point $B,$ calculate coordinates of point $C$ on red curve. Using $x$ coordinate of point $C,$ calculate coordinates of point $D$ on black curve. Points $A,B,C,D$ enclose the point of intersection of the 2 curves.

Point of intersection

Graphs of 2 curves showing complex cube root enclosed by four points $A,B,C,D$. Point $E,$ intersection of lines $AC,BD$ is close to complex cube root, and is starting point for next iteration. Calculate equations of lines $AC,BD.$ Calculate coordinates of point $E,$ intersection of lines $AC,BD.$ Point $E$ is used as starting point for next iteration. Area of quadrilateral $ABCD$ becomes smaller and smaller until complex cube root, intersection of red and black curves, is identified.

Implementation

Initialization

# python code import decimal D = decimal.Decimal precision = decimal.getcontext().prec = 100 useDecimal = 1 Tolerance = 1e-14 if useDecimal : Tolerance = D('1e-' + str(decimal.getcontext().prec - 2)) def line_mc (point1,point2) : ''' m,c = line_mc (point1,point2) where y = mx + c. ''' x1,y1 = point1 x2,y2 = point2 m = (y2-y1) / (x2-x1) # y = mx + c c = y1 - m*x1 return m,c def intersectionOf2Lines (line1, line2) : m1,c1 = line1 m2,c2 = line2 # y = m1x + c1 # y = m2x + c2 # m1x + c1 = m2x + c2 # m1x - m2x = c2 - c1 x = (c2-c1)/(m1-m2) y = m1*x + c1 return x,y def almostEqual (v1,v2,tolerance = Tolerance) : ''' status = almostEqual (v1,v2) For floats, tolerance is 1e-14. 1234567.8901234567 and 1234567.8901234893 are not almostEqual. 1234567.8901234567 and 1234567.8901234593 are almostEqual. ''' return abs(v1-v2) < tolerance*abs((v1+v2)/2)

Function two_points()

Graphs of 2 curves showing 4 points that enclose one of the complex cube roots. On first invocation function two_points() returns points $A,B.$ On second invocation function two_points() returns points $C,D.$ If distance $AB$ or distance $CD$ is very small, function two_points() returns status True. def two_points (y,a,b,quadrant) : ''' [point1,point2],status = two_points (y,a,b) ''' L1 = [] ; yInput = y if 0 : print ('two_points():') s1 = ' a,b,y' ; print (s1, eval(s1)) # a = ppp - pqq - 2pqq = ppp - 3pqq (1) # b = 2ppq + ppq - qqq = 3ppq - qqq (2) # # Using (2) # 3xxy - yyy = b # # yyy + b # X = ---------- # 3y X = (y**3 + b) / (3*y) if isinstance(X,D) : x = X.sqrt() else : x = X ** 0.5 if quadrant in (2,3) : x *= -1 L1 += [(x,y)] # Using (1) # xxx - 3xyy = a # # xxx - a # Y = ----------- # 3x # Y = (x**3 - a) / (3*x) if isinstance(Y,D) : y = Y.sqrt() else : y = Y ** 0.5 if quadrant in (3,4) : y *= -1 L1 += [(x,y)] return L1, almostEqual(y, yInput)

Function pointOfIntersection()

Graphs of 2 curves showing complex cube root enclosed by four points $A,B,C,D$. From points $A,B,C,D$ function pointOfIntersection() calculates coordinates of point $E.$ If distance $AB$ is very small, point $A$ is returned as equivalent to intersection of red and black curves. If distance $CD$ is very small, point $C$ is returned as equivalent to intersection of red and black curves. def pointOfIntersection(y,a,b,quadrant) : ''' pointE, status = pointOfIntersection(y,a,b) y is Y coordinate of point A. ''' # print('\npointOfIntersection()') t1,status = two_points (y,a,b,quadrant) ptA,ptB = t1 if status : # Distance between ptA and ptB is very small. # ptA is considered equivalent to the complex cube root. return ptA,status t2,status = two_points (ptB[1],a,b,quadrant) ptC,ptD = t2 if status : # Distance between ptC and ptD is very small. # ptC is considered equivalent to the complex cube root. return ptC,status lineAC = line_mc (ptA,ptC) lineBD = line_mc (ptB,ptD) pointE = intersectionOf2Lines (lineAC, lineBD) return pointE,False

Execution

def CheckMake_pq(a,b,x,y) : ''' p,q = CheckMake_pq(a,b,x,y) p = x and q = y. p,q are checked as valid within tolerance, and are reformatted slightly to improve appearance. ''' # print ('\nCheckMake_pq(a,b,x,y)') # s1 = '(a,b)' ; print (s1,eval(s1)) # s1 = ' x'  ; print (s1,eval(s1)) # s1 = ' y'  ; print (s1,eval(s1)) P = x**2 ; Q = y**2 ; p=q=-1 for p in (x,) : # a = ppp - 3pqq; a + 3pqq should equal ppp. if not almostEqual (P*p, a + 3*p*Q, Tolerance*100) : continue for q in (y,) : # b = 3ppq - qqq; b + qqq should equal 3ppq if not almostEqual (3*P*q, b + q*Q) : continue # Following 2 lines improve appearance of p,q if useDecimal : # 293.00000000000000000000000000000000000000034 becomes 293 p,q = [ decimal.Context(prec=precision-3).create_decimal(s).normalize() for s in (p,q) ] else : # 123.99999999999923 becomes 124.0 p,q = [ float(decimal.Context(prec=14).create_decimal(s)) for s in (p,q) ] return p,q # If code gets to here there is internal error. s1 = ' p' ; print (s1,eval(s1)) s1 = ' q' ; print (s1,eval(s1)) s1 = 'P*p, a + 3*p*Q' ; print (s1,eval(s1)) s1 = '3*P*q, b + q*Q' ; print (s1,eval(s1)) 1/0 def ComplexCubeRoot (a,b, y = 100, count_ = 20) : ''' p,q = ComplexCubeRoot (a,b, y, count_) (p+qi)**3 = (a+bi) ''' print ('\nComplexCubeRoot(): a,b,y,count_ =',a,b,y,count_) if useDecimal : y,a,b = [ D(str(v)) for v in (y,a,b) ] if a == 0 : if b == 0 : return 0,0 if useDecimal : cubeRoot = abs(b) ** (D(1)/3) else : cubeRoot = abs(b) ** (1/3) if b > 0 : return 0,-cubeRoot return 0,cubeRoot if b == 0 : if useDecimal : cubeRoot = abs(a) ** (D(1)/3) else : cubeRoot = abs(a) ** (1/3) if a > 0 : return cubeRoot,0 return -cubeRoot,0 # Select most appropriate quadrant. if a > 0: setx = {2,3} else: setx = {1,4} if b > 0: sety = {1,2} else: sety = {3,4} quadrant, = setx & sety # Make sign of y appropriate for this quadrant. if quadrant in (1,2) : y = abs(y) else : y = -abs(y) s1 = ' quadrant' ; print (s1, eval(s1)) for count in range (0,count_) : pointE,status = pointOfIntersection(y,a,b,quadrant) s1 = 'count,status' ; print (s1, eval(s1)) s1 = ' pointE[0]' ; print (s1, eval(s1)) s1 = ' pointE[1]' ; print (s1, eval(s1)) x,y = pointE if status : break p,q = CheckMake_pq(a,b,x,y) return p,q

An Example

p,q = 18,-29 w0 = p + q*1j W = w0**3 a,b = W.real, W.imag s1 = '\na,b' ; print (s1, eval(s1)) print ('Calculate one cube root of W =', W) p,q = ComplexCubeRoot (a, b, -18) s1 = '\np,q' ; print (s1, eval(s1)) sign = ' + ' if q < 0 : sign = ' - ' print ('w0 = ', str(p) ,sign, str(abs(q)),'i',sep='') a,b (-39582.0, -3799.0) Calculate one cube root of W = (-39582-3799j) ComplexCubeRoot(): a,b,y,count_ = -39582.0 -3799.0 -18 20 quadrant 4 count,status (0, False) pointE[0] 18.29530595866769796981147594954794157453427770441979517949705190002312860122683512802090262517713985 pointE[1] -29.17605851188829785804056660826030733025475591125914094664311767722817017040959484906193571713185723 count,status (1, False) pointE[0] 18.00005338608833140244623119091867294731079673210031643698475740639013316464725974710029414039192178 pointE[1] -29.00004871113589733281025047965490410760310240487028781197782902118493613318468122514940700337153822 count,status (2, False) pointE[0] 18.00000000000411724901243622639913339568271402799883998577879934397861645683113192688877413853607310 pointE[1] -29.00000000000374389488957142528693701977412078931714190614174861872117809632376941851192785888688849 count,status (3, False) pointE[0] 18.00000000000000000000000002432197332441306371168312765664226520285586754485200564671348858119116700 pointE[1] -29.00000000000000000000000002211642401405406173668734741733917293863102998696594080783163691859719835 count,status (4, False) pointE[0] 18.00000000000000000000000000000000000000000000000000000084875301166228708732099292145747224660296019 pointE[1] -29.00000000000000000000000000000000000000000000000000000077178694502906372553368739653233322951192943 count,status (5, False) pointE[0] 18.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 pointE[1] -29.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 count,status (6, True) pointE[0] 18 pointE[1] -29.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 p,q (Decimal('18'), Decimal('-29')) w0 = 18 - 29i

Method #3 (Algebraic)

Introduction

Let complex number $w=p+qi.$

Then $W=w^{3}=(p+qi)^{3}=p^{3}-3pq^{2}+(3p^{2}q-q^{3})i.$

Let $W=a+bi.$

Then:

$a=p^{3}-3pq^{2}\ \dots \ (1)$ and

$b=3p^{2}q-q^{3}\ \dots \ (2).$

When $W$ is given and $w$ is desired, $w$ may be calculated from the solutions of 2 simultaneous equations $(1)$ and $(2),$ where $a,b$ are known values, and $p,q$ are desired.

Implementation

$(1)$ squared: $a^{2}=p^{6}-6p^{4}q^{2}\ +9p^{2}q^{4}\dots \ (1a)$

From $(2):\ 3p^{2}q=b+q^{3}\ \dots \ (2a)$

$(1a)*27q^{3}:$

$27q^{3}a^{2}=27q^{3}p^{6}-27q^{3}(6)p^{4}q^{2}\ +27q^{3}(9)p^{2}q^{4}$

$27q^{3}a^{2}=27(p^{2}q)^{3}-27(6)p^{4}q^{5}\ +27(9)p^{2}q^{7}$

$27q^{3}a^{2}=(3p^{2}q)^{3}-27(6)p^{4}q^{2}q^{3}\ +27(9)p^{2}qq^{6}$

$27q^{3}a^{2}=(3p^{2}q)^{3}-3(6)(9p^{4}q^{2})q^{3}\ +27(3)(3p^{2}q)q^{6}$

$27q^{3}a^{2}=(3p^{2}q)^{3}-3(6)((3p^{2}q)^{2})q^{3}\ +27(3)(3p^{2}q)q^{6}$

Let $Q=q^{3}:$

$27Qa^{2}=(3p^{2}q)^{3}-3(6)((3p^{2}q)^{2})Q+27(3)(3p^{2}q)Q^{2}\ \dots \ (1b)$

For $(3p^{2}q)$ in $(1b)$ substitute $(b+Q):$

$27Qa^{2}=(b+Q)^{3}-3(6)((b+Q)^{2})Q+27(3)(b+Q)Q^{2}\ \dots \ (1c)$

Expand $(1c),$ simplify, gather like terms and result is:

$f(Q)=sQ^{3}+tQ^{2}+uQ+v\ \dots \ (3)$ where:

$Q=q^{3}$

$s=64$

$t=48b$

$u=-(15b^{2}+27a^{2})$

$v=b^{3}$

Calculate one real root of $(3):\ Q_{1}$

$q_{1}={\sqrt[{3}]{Q_{1}}}$

From $(2a):\ p_{1}={\sqrt {\frac {b+Q_{1}}{3q_{1}}}}$

Check $p_{1}$ against $(1)$ to resolve ambiguity of sign of $p_{1}.$

$p_{1}+q_{1}i$ is one cube root of $W.$

$p$ may be calculated without ambiguity as follows: $a=p^{3}-3pq^{2}\ \dots \ (1)$ and $b=3p^{2}q-q^{3}\ \dots \ (2).$ From $(1):\ p^{3}-3q^{2}p-a=0\ \dots \ (3)$ From $(2):\ 3qp^{2}-(Q+b)=0\ \dots \ (4)$ Let: $A=-3q^{2}$ $B=-a$ $C=3q$ $D=-(Q+b)$ Then $(3),(4)$ become: $p^{3}+Ap+B=0\ \dots \ (5)$ $Cp^{2}+D=0\ \dots \ (6)$ $(5)*D:\ Dp^{3}+DAp+DB=0\ \dots \ (7)$ $(6)*B:\ BCp^{2}+BD=0\ \dots \ (8)$ $(7)-(8):\ Dp^{3}-BCp^{2}+DAp=0\ \dots \ (9)$ Simplify $(9):\ Dp^{2}-BCp+DA=0\ \dots \ (10)$ Repeat $(6):\ Cp^{2}+D=0\ \dots \ (6)$ $(10)*C:\ CDp^{2}-BCCp+DAC=0\ \dots \ (11)$ $(6)*D:\ CDp^{2}+DD=0\ \dots \ (12)$ $(11)-(12):\ -BCCp+DAC-DD=0\ \dots \ (13)$ From $(13):\ p={\frac {DAC-D^{2}}{BC^{2}}}={\frac {D(AC-D)}{BC^{2}}}$

An Example

Calculate cube roots of complex number $W=39582+3799i.$

Graph of $f(Q)$ shown as graph of $f(x)$ and showing three values of $Q:Q_{1},Q_{2},Q_{3}$. $Y$ axis compressed for clarity. # python code: a,b = 39582,3799 s = 64 t = 48*b u = -(15*b**2 + 27*a**2) v = b**3 s,t,u,v (64, 182352, -42518323563, 54828691399) Calculate roots of cubic function: $y=f(x)$$=64x^{3}$$+182352x^{2}$$-42518323563x$$+54828691399.$ Three roots are: $Q_{1},Q_{2},Q_{3}=-27239.53953801976,1.2895380197588122,24389$

# python code:

values_of_Q = Q1,Q2,Q3 = -27239.53953801976, 1.2895380197588122, 24389
q1,q2,q3 = [ abs(Q)**(1/3) for Q in values_of_Q ]
q1 *= -1
values_of_q = q1,q2,q3
s1 = 'values_of_q' ; print(s1, eval(s1))

for m in zip(values_of_q, values_of_Q) :
    q,Q = m
    p = ((b + Q)/(3*q)) ** .5
    sum = p**3 - 3*p*q**2 - a
    if abs(sum) > 1e-10 : p *= -1
    w = p + q*1j
    s1 = 'w, w**3' ; print (s1, eval(s1))

values_of_q (-30.08845726811989, 1.0884572681198943, 29)
w, w**3 ((-16.114736709748723 - 30.08845726811989j  ), (39582+3799j))
w, w**3 (( 34.11473670974874  +  1.0884572681198943j), (39582+3799j))
w, w**3 ((-18.0               + 29.0j               ), (39582+3799j))

Three cube roots of $W=39582+3799i$ are:

$w_{1}=(-16.114736709748723-30.08845726811989i)$

$w_{2}=(34.11473670974874+1.0884572681198943i)$

$w_{3}=(-18.0+29.0i)$