In general, any function F(z) could return a complex value, unless F(z) is defined to be purely real or purely imaginary. If variable z is allowed to be complex, then F(z) can be expressed as a sum of real and imaginary terms:
Functions Fr and Fi can be found by substituting (z = x + iy) or (z = R eiq) into F(z), and then simplifying the results until F(z) becomes a sum of a real part and imaginary part. For example, suppose that F(z) is a power of z,
where z is a complex variable, and c and k are real-valued constants. F(z) can be resolved by substituting the complex-polar form of z,
and then replacing eikq with Cos(kq) + iSin(kq):
c zk = c Rk Cos(kq) + ic Rk Sin(kq)
The last expression is a sum of a real term and an imaginary term, which corresponds to the sum
Fi = c Rk Sin(kq)
Here are some guidelines on which form of z to substitute in F(z):
Term in F(z) | Fr and Fi | Example |
Power czk
z = R eiq, c and k are real (analysis) | Fr = c Rk Cos(kq)
Fi = c Rk Sin(kq) | 5 z3 = 5R3 Cos(3q) + i5R3 Sin(3q) |
Power czk
c = p eif, z = R eiq, k = m + in (analysis) | Fr = p Rm e-nq Cos[n Ln(R) + mq + f]
Fi = p Rm e-nq Sin[n Ln(R) + mq + f] | 4 e3i z(1 - 2i)
= 4R e2q Cos[-2 Ln(R) + q + 3] + i4R e2q Sin[-2Ln(R) + q + 3] |
Exponential Function cekz
c = p eif, k = m + in, z = x + iy (analysis) | Fr = p Exp[(mx - ny)] Cos(my + nx + f)
Fi = p Exp[(mx - ny)] Sin(my + nx + f) | 4 e3i Exp[(1 - 2i)(x + iy)]
= 4ex + 2y Cos(y - 2x + 3) + i4ex + 2y Sin(y - 2x + 3) |
Natural Logarithm Ln(cz)
c = p eif, z = R eiq (analysis) | Fr = Ln (pR)
Fi = q + f | Ln[4 e3iz] = Ln(4R) + i(q + 3) |
Trig Function Cos(kz)
k = m + in, z = x + iy (analysis) | Fr = Cosh(nx + my) Cos(mx - ny)
Fi = - Sinh(nx + my) Sin(mx - ny) | Cos[(2 - 3i)z]
= Cosh(-3x + 2y) Cos(2x + 3y) - iSinh(-3x + 2y) Sin(2x + 3y) |
Trig Function Sin(kz)
k = m + in, z = x + iy (analysis) | Fr = Cosh(nx + my) Sin(mx - ny)
Fi = Sinh(nx + my) Cos(mx - ny) | Sin[(4 - 7i)]
= Cosh(-7x + 4y) Sin(4x + 7y) + iSinh(-7x + 4y) Cos(4x + 7y) |
Resolve c zk by substituting the complex-polar form of z:
c zk = c Rk eikq
c zk = c Rk [Cos(kq) + iSin(kq)]
This simplifies to
Let c = p eif, z = R eiq and k = m + in:
c zk = p eif Rm + in eiq(m + in)
c zk = p eif Rm Rin eimq e-nq
Move the factors with real-exponents to the left, and those with imaginary-exponents to the right:
Replace Rin with ein Ln(R), then combine all of the imaginary exponents:
c zk = p Rm e-nq ei[n Ln(R) + mq + f]
Substitute
Let c = p eif, k = m + in, and z = x + iy:
c ekz = p eif emx - ny + imy + inx
c ekz = p emx - ny ei(my + nx + f)
This simplifies to
Let c = p eif and z = R eiq:
Ln (cz) = Ln [p R ei(q + f)]
Ln (cz) = Ln [p R] + Ln [ei(q + f)]
This simplifies to
Let k = m + in and z = x + iy:
Cos(kz) = Cos[(mx - ny) + i(nx + my)]
Cos(kz) = ½ [ei[(mx - ny) + i(nx + my)] + e-i[(mx - ny) + i(nx + my)] ]
Cos(kz) = ½ [e-(nx + my) ei(mx - ny)] + ½ [e(nx + my) e-i(mx - ny) ]
Cos(kz) = ½ e-(nx + my) [Cos(mx - ny) + iSin(mx - ny)] + ½ enx + my [Cos(mx - ny) - iSin(mx - ny)]
Cos(kz) = ½ [enx + my + e-(nx + my) ][Cos(mx - ny)] - i½ [e+(nx + my) - e-(nx + my) ][Sin(mx - ny)]
The Hyperbolic Cosine and Sine are defined
Let k = m + in and z = x + iy:
Sin(kz) = Sin [(mx - ny) + i(nx + my)]
Sin(kz) = -½i[ei[(mx - ny) + i(nx + my)] - e-i[(mx - ny) + i(nx + my)] ]
Sin(kz) = -½i[e-(nx + my) ei(mx - ny) ] + ½i[e(nx + my) e-i(mx - ny) ]
Sin(kz) = -½ie-(nx + my) [Cos(mx - ny) + iSin(mx - ny)] + ½ie(nx + my) [Cos(mx - ny) - iSin(mx - ny)]
Sin(kz) = ½ [e(nx + my) + e-(nx + my) ][Sin(mx - ny)] + i½ [e+(nx + my) - e-(nx + my) ][Cos(mx - ny)]
Substitution of the Hyperbolic Sine and Cosine yields a simplified expression:
URL: http://members.aceweb.com/patrussell/approximations/Resolve.htm
Unpublished Work. © Copyright 2001 Pat Russell. Updated April 17, 2009.