Index

# Resolving Functions Into Real and Imaginary Parts

## 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 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,

F(z) = c zk

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 zk = c (R eiq)k = c Rk eikq

and then replacing eikq with Cos(kq) + iSin(kq):

c zk = c Rk [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 Fr + iFi. Thus, we can write

Fr = c Rk Cos(kq)

Fi = c Rk 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 eiq.
• If z is the argument of a logarithm, then substitute z = R eiq.
• If constant-k and variable-z are both complex, and they are multiplied together (i.e. k*z), then substitute k = r eif and z = R eiq. The result will be rR*ei(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 eiq) at the Complex Numbers page.

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

 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)

## Detailed Analysis of Resolved Functions

### Power czk where c and k are real

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

c zk = (R eiq)k

c zk = c Rk eikq

c zk = c Rk [Cos(kq) + iSin(kq)]

This simplifies to

c zk = c Rk Cos(kq) + ic Rk Sin(kq)

### Power czk when c, k and z are complex

Let c = p eif, z = R eiq and k = m + in:

c zk = (R eiq)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:

c zk = p Rm e-nq Rin eif eimq

Replace Rin with ein Ln(R), then combine all of the imaginary exponents:

c zk = Rm e-nq ein Ln(R) eimq eif

c zk = p Rm e-nq ei[n Ln(R) + mq + f]

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

c zk = p Rm e-nq Cos[nLn(R) + mq + f] + ip Rm e-nq Sin[n Ln(R) + mq + f]

### Exponential function c ekz when c, k and z are complex

Let c = p eif, k = m + in, and z = x + iy:

c ekz = p eif e(m + in)(x + iy)

c ekz = p eif emx - ny + imy + inx

c ekz = p emx - ny ei(my + nx + f)

This simplifies to

c ekz = p emx - ny Cos(my + nx + f) + ip emx - ny Sin(my + nx + f)

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

Let c = p eif and z = R eiq:

Ln (cz) = Ln [p eif R eiq]

Ln (cz) = Ln [p R ei(q + f)]

Ln (cz) = Ln [p R] + Ln [ei(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 [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 Cosh(u) = ½ (eu + e-u ) and Sinh(u) = ½ (eu - 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[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:

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

URL: http://members.aceweb.com/patrussell/approximations/Resolve.htm
Unpublished Work. © Copyright 2001 Pat Russell. Updated April 17, 2009.