3. ELEMENTARY
         FUNCTIONS    
    

Exponential and Trigonometric Functions

If z = x + iy, we define the exponential function exp z = ex cis y (= ez).

Notes

1. We can give an alternative definition in terms of power series.
Writing out a formal series for e iy gives cis y.

2. If y = 0, then exp z = exp x = ex. Thus the complex exponential function naturally extends the real function.

3. In this definition, y is in radian measure.

Properties of the Exponential (I)

1. The function exp is entire and (exp z) = exp z. (See Thm 2.5 and next example.)

2. If w = w(z) is analytic in some domain D, then so is exp w.

3. The function exp z = ex cis y is a complex number in polar form:
                       |exp z| = ex,  arg(exp z) = y.

4. The range of w = exp z is the whole w-plane except O.
For, w = ex cis y with ex > 0; to get w = cis , set x = ln ( > 0) and y = .

5. Laws of exponents
                       exp z1 . exp z2 = exp(z1 + z2)
                       exp z1 / exp z2 = exp(z1 – z2)

6. Powers
                        (exp z)m = exp(mz)  m Z+
                       (exp z)1/n = exp 1/n (z + 2ki)  m, n Z+
                       (exp z)m/n = exp m/n (z + 2ki)  k Z.

The proofs follow directly from the definition of the exponential.
Note that
                     (ex cis y)1/n = ex/n cis ((y + 2k)/n).

Properties of the Exponential (II)

8. We observe that  exp(z + 2i) = exp z.exp(2i),

and that  exp(2i) = e0.(cos 2 + i sin 2) = 1.

It follows that exp(z + 2i) = exp z.

Thus we can divide the z-plane into periodic strips. Each strip in the z-plane is mapped to the whole w-plane excluding the origin.

We note the further two properties of the exponential:

8. exp = .
9. cis = cos + i sin = exp(i).


Quiz 3.1     


Sine and Cosine

If y is a real number, we have

  exp(iy) = cos y + i sin y,
exp(–iy) = cos y – i sin y,

and so

cos y = 1/2. (exp(iy) + exp(–iy)),

sin y = 1/2i.(exp(iy) – exp(–iy)).

Thus it is natural to define cosine and sine as:

cos z = 1/2. (exp(iz) + exp(–iz)),

sin z = 1/2i.(exp(iz) – exp(–iz)).

These are Euler's relations. Again notice here how we try to generalize, or extend, a ‘real’ situation to the complex case.

Properties of Sine and Cosine

1.  Both functions are entire:

(sin z) = cos z, (cos z) = – sin z.

2.  Both functions are periodic, of period 2. This follows from the periodicity of the exponential function.

The functions satisfy the usual identities, as in the real case.

3. sin2 z + cos2 z = 1.

4.  sin(z1 + z2) = sin z1 cos z2 + sin z2 cos z1  etc.

5.  sin(– z) = – sin z, cos(– z) = cos z etc.


Logarithmic Function

Does the exponential function have an inverse logarithmic function? Since the exponential function is periodic, any inverse would have to be multi-valued. Let us write

w = log z z = exp w.

If we set z = r cis , w = u + iv, then r cis = eu cis v.

From this, we deduce that

r = eu, u = ln r, v = + 2k.

That is,

w = log z = ln | z | + i( + 2k) (k Z).

Thus there are infinitely many values of log z, the different values differing by 2ki. Each value of k gives a branch of the logarithm.


The Cut Plane

With log z = ln | z | + i( + 2k) let us take – < .

Make a (red) cut in the complex plane along the negative x-axis. For any fixed value of k, we obtain a branch which does not cross this cut. So in the cut plane, each branch is single-valued. In particular we have the principal branch

Log z = ln r + i   (– < ).

Notes

1. A path which crosses the cut moves to the next branch.

2. If z is real and positive, then Log z = ln r.

3. We can think of the branch planes interleaved together, with the x-axis as a common axis. A path drawn about the origin in one branch plane reaches the cut and then passes to the next branch plane.

4. Our choice of the positive x-axis for the cut was somewhat arbitrary. Other branch cuts are possible; but O is common to them all – O is a branch point.


Properties of the Logarithm (I)

Consider Log z = ln r + i   (– < , r > 0) – that is, over the open domain excluding the cut. There are difficulties on the cut, for is not continuous there for any branch. Hence, for example, the Log function is not continuous on the cut, and so the Log function is not differentiable there.

1. Log z is analytic over the open domain (– < < , r > 0).

Writing Log z = u + iv, we have u = 1/2 ln (x2 + y2),  v = = arctan y/x.
Hence

These functions are continuous on the given domain and satisfy the Cauchy-Riemann equations there. Hence by Theorem 2.5, Log z is analytic.

[Note There is a problem in defining arctan here when x = 0. We could overcome this by defining = arccot x/y, or by taking time to develop a polar form of the Cauchy-Riemann equations.]

Properties of the Logarithm (II)

2.  Derivative

All branches have the same derivative, since they differ by a constant.

3. Inverse Property

exp(log z) = z (for any branch)      
log(exp z) = z (for a particular branch).

4. Sums and Differences

log z1 + log z2 = log(z1 . z2)    
log z1 – log z2 = log(z1 / z2)    

providing we choose the appropriate logarithm branch on the right.

Examples on the Logarithm

Example 1. Evaluate Log(–1) + Log(–1).

Now –1 = 1 . cis , so Log(–1) = 0 + i.

Hence 2Log(–1) = 2i = log 1, but not Log 1 (= 0).

Example 2. Show how to make f(z)=log z analytic on the open region A = G R.

In (green) region G, we define f(z) = Log z (the principal value).
In (red) region R, we choose a different branch of the logarithm, defining
f(z) = log | z | + i arg z ( < arg z < 3).

This definition allows a continuous transition acoss the cut.




Complex Exponents

Using our knowledge of real powers, we define the complex power zc (c C) by

zc = exp(c log z), (z 0).

Since zc is defined in terms of the logarithm, we expect zc to be multivalued, so we use the cut plane as for the logarithm. Then since log z is single-valued and analytic in the cut plane, so is zc. Now

So

Exponent Examples

1.   i 1/4 = exp(1/4 log i) = exp(1/4 i (/2 2k)) = exp(i/8 ki/2) – four values.

2. i i = exp(i log i) = exp(i (/2 2k) i) = exp(–/2 2k).
The principal value is exp(–/2).

3. What is the relationship between exp z and ez ?

Clearly ez = exp(z log e). Now e = e cis 0, so log e = 1 2ki, and
exp(z log e) = exp(z 2kiz).

It follows that ez = exp z . exp(2kiz). Setting k = 0 gives ez = exp z.
Thus exp z is the principal value of the multi-valued power function ez.

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