Resolving Functions Into Real and Imaginary Parts

What "resolving" functions means
Table of Resolved Complex Functions
Detailed Analysis of Resolved Functions

What "resolving" functions means

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:

F(z) = Fr + iFi

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,

F(z) = c z^k

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,

c z^k = c (R e^iq)^k = c R^k e^ikq

and then replacing e^ikq with Cos(kq) + iSin(kq):

c z^k = c R^k [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 Fr + iFi. Thus, we can write

Fr = c R^k Cos(kq)

Fi = c R^k Sin(kq)

Here are some guidelines on which form of z to substitute in F(z):

If z is raised to exponent N, and N does not equal one or zero, then substitute z = R e^iq.
If z is the argument of a logarithm, then substitute z = R e^iq.
If constant-k and variable-z are both complex, and they are multiplied together (i.e. k*z), then substitute k = r e^if and z = R e^iq. The result will be rR*e^{i(f + q)}. If this result has no other complex-coefficients, then it resolves to rR*Cos(f + q) + irR*Sin(f + q).
In most other cases, substitute z = x + iy.
If a fraction has a complex denominator, then multiply the numerator and denominator by the denominator's complex conjugate. The result is a complex-fraction with a real-valued denominator.
Remember that dividing by imaginary number i is equivalent to multiplying by -i.
If needed, review the formulas relating complex-linear expressions (z = x + iy) and complex-polar (z = R e^iq) at the Complex Numbers page.

Table of Resolved Complex Functions, F(z) = Fr + iFi

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 z³ = 5R³ Cos(3q) + i5R³ 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 e³ⁱ 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 e³ⁱ 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 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)

Detailed Analysis of Resolved Functions

Power cz^k where c and k are real

Resolve c z^k by substituting the complex-polar form of z:

c z^k = (R e^iq)^k

c z^k = c R^k e^ikq

c z^k = c R^k [Cos(kq) + iSin(kq)]

This simplifies to

c z^k = c R^k Cos(kq) + ic R^k Sin(kq)

Power cz^k when c, k and z are complex

Let c = p e^if, z = R e^iq and k = m + in:

c z^k = (R e^iq)^{m + in}

c z^k = p e^if R^{m + in} e^{iq(m + in)}

c z^k = p e^if R^m Rⁱⁿ e^imq e^-nq

Move the factors with real-exponents to the left, and those with imaginary-exponents to the right:

c z^k = p R^m e^-nq Rⁱⁿ e^if e^imq

Replace Rⁱⁿ with e^{in Ln(R)}, then combine all of the imaginary exponents:

c z^k = R^m e^-nq e^{in Ln(R)} e^imq e^if

c z^k = p R^m e^-nq e^{i[n Ln(R) + mq + f]}

Substitute Cos[nLn(R) + mq + f] + iSin[nLn(R) + mq + f] for the equivalent exponential term to finish:

c z^k = p R^m e^-nq Cos[nLn(R) + mq + f] + ip R^m e^-nq Sin[n Ln(R) + mq + f]

Exponential function c e^kz when c, k and z are complex

Let c = p e^if, k = m + in, and z = x + iy:

c e^kz = p e^if e^{(m + in)(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

c e^kz = p e^{mx - ny} Cos(my + nx + f) + ip e^{mx - ny} Sin(my + nx + f)

Natural logarithm Ln(cz) when c and z are complex

Let c = p e^if and z = R e^iq:

Ln (cz) = Ln [p e^if 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

Ln (cz) = Ln (pR) + i(q + f)

Trigonometric term Cos(kz) when k and z are complex

Let k = m + in and z = x + iy:

Cos(kz) = Cos[(m + in)(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 Cosh(u) = ½ (e^u + e^-u) and Sinh(u) = ½ (e^u - e^-u) . Substituting Sinh and Cosh in the above equation gives a simplified expression:

Cos(kz) = Cosh(nx + my) Cos(mx - ny) - iSinh(nx + my) Sin(mx - ny)

Trigonometric term Sin(kz) when k and z are complex

Let k = m + in and z = x + iy:

Sin(kz) = Sin [(m + in)(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:

Sin(kz) = Cosh(nx + my) Sin(mx - ny) + iSinh(nx + my) Cos(mx - ny)

Index

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 z³ = 5R³ Cos(3q) + i5R³ 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 e³ⁱ 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 e³ⁱ 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 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)