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 e^{iq}) 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 e^{ikq} with Cos(kq) + iSin(kq):
c z^{k} = c R^{k} Cos(kq) + ic R^{k} Sin(kq)
The last expression is a sum of a real term and an imaginary term, which corresponds to the sum
Fi = c R^{k} 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 cz^{k}
z = R e^{iq}, c and k are real (analysis) | Fr = c R^{k} Cos(kq)
Fi = c R^{k} Sin(kq) | 5 z^{3} = 5R^{3} Cos(3q) + i5R^{3} Sin(3q) |
Power cz^{k}
c = p e^{if}, z = R e^{iq}, k = m + in (analysis) | Fr = p R^{m} e^{-nq} Cos[n Ln(R) + mq + f]
Fi = p R^{m} e^{-nq} Sin[n Ln(R) + mq + f] | 4 e^{3i} z^{(1 - 2i)}
= 4R e^{2q} Cos[-2 Ln(R) + q + 3] + i4R e^{2q} Sin[-2Ln(R) + q + 3] |
Exponential Function ce^{kz}
c = p e^{if}, 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 e^{3i} Exp[(1 - 2i)(x + iy)]
= 4e^{x + 2y} Cos(y - 2x + 3) + i4e^{x + 2y} Sin(y - 2x + 3) |
Natural Logarithm Ln(cz)
c = p e^{if}, z = R e^{iq} (analysis) | Fr = Ln (pR)
Fi = q + f | Ln[4 e^{3i}z] = 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 z^{k} by substituting the complex-polar form of z:
c z^{k} = c R^{k} e^{ikq}
c z^{k} = c R^{k} [Cos(kq) + iSin(kq)]
This simplifies to
Let c = p e^{if}, z = R e^{iq} and k = m + in:
c z^{k} = p e^{if} R^{m + in} e^{iq(m + in)}
c z^{k} = p e^{if} R^{m} R^{in} e^{imq} e^{-nq}
Move the factors with real-exponents to the left, and those with imaginary-exponents to the right:
Replace R^{in} with e^{in Ln(R)}, then combine all of the imaginary exponents:
c z^{k} = p R^{m} e^{-nq} e^{i[n Ln(R) + mq + f]}
Substitute
Let c = p e^{if}, k = m + in, and z = x + iy:
c e^{kz} = p e^{if} e^{mx - ny + imy + inx}
c e^{kz} = p e^{mx - ny} e^{i(my + nx + f)}
This simplifies to
Let c = p e^{if} and z = R e^{iq}:
Ln (cz) = Ln [p R e^{i(q + f)}]
Ln (cz) = Ln [p R] + Ln [e^{i(q + f)}]
This simplifies to
Let k = m + in and z = x + iy:
Cos(kz) = Cos[(mx - ny) + i(nx + my)]
Cos(kz) = ½ [e^{i[(mx - ny) + i(nx + my)] } + e^{-i[(mx - ny) + i(nx + my)] }]
Cos(kz) = ½ [e^{-(nx + my)} e^{i(mx - ny)}] + ½ [e^{(nx + my)} e^{-i(mx - ny) } ]
Cos(kz) = ½ e^{-(nx + my) } [Cos(mx - ny) + iSin(mx - ny)] + ½ e^{nx + my} [Cos(mx - ny) - iSin(mx - ny)]
Cos(kz) = ½ [e^{nx + 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[e^{i[(mx - ny) + i(nx + my)] } - e^{-i[(mx - ny) + i(nx + my)] }]
Sin(kz) = -½i[e^{-(nx + my) } e^{i(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.