Coursera: Analysis of a Complex Kind. 4.1 Inverse analytic functions

Jul 24, 2016

say z = e^w \, w \in \, C  \backslash 0, (i) then what is the function, L, that maps z to w? That is L(z) = w?

Say our “starting” value, z = |z|e^{i \theta} (ii)

Consider first of all equating (i) and (ii) z = e^w \, = \, |z| e^{i \theta}

Now write w = u + iv

So, (z =) e^{u + iv} \, = e^{u} e^{iv} \, = \, |z| e^{i \theta}

By equating real and complex parts (??):

e^u \, = \, |z| \,  ,e^{iv} \, = e^{i \theta}

So u = ln |z|, and v = θ + 2kπ = arg z (from definition of exponential function).

So Log(z) = w = u+iv = ln|z| + i (θ) =  ln|z| + i Arg (z), principal argument

and more generally

log(z) = w = u+iv = ln|z| + i (θ + 2kπ) =  ln|z| + i arg (z), multi-valued argument

Slide 3 Example values of Log function

Note , function of a complex variable is defined as “combination” of two functions of real variables.

Log(1) = ln |1| + i Arg(1) = 0 + i(0) = 0

Log(i) = ln|i| + i Arg(i) = ln(1) + i(π/2) = i(π/2)

Log(-1) = ln|-1| + i Arg(-1) = ln(1) + i(π) = iπ

Log (1 + i) = ln | 1 + i| + i Arg(1 + i) = √2 + i(π/4)

so the definition of the log function has been extended to negative and complex numbers except still undefined at zero.

Slide 4 Continuity of the Log function

Log(z) = ln |z| + i Arg z

z -> |z| is continuous in C

z -> ln|z|  is continuous in C \{0}

z -> Arg z is continuous in C\(-\infty, 0]

Thus Log z is continuous in C\(-\infty, 0]

However, as z -> -x  \; \in (-\infty, 0) (*), from above , Log(z) -> ln |z| +

and, since z -> -x  \; \in (-\infty, 0) (*), from below , Log(z) -> ln |z|  iπ

(* z is getting close to the negative real axis)

So Log (z) is not continuous on (-\infty,0), and not defined at o.

Slide 5 Is the Log function analytic ?

Result: The principal branch of the Log function is analytic in C\(-\infty,0)

Derivative of Log z

e^{Log \; z} = z

Differentiating by chain rule

e^{Log \; z}.\frac{d}{dz}(Log \; z) = 1

So, \frac{d}{dz}(Log \; z) = \frac{1}{z}

Slide 6 Differentiating inverse functions

slide 4.1.6

f is analytic and maps the region (domain) U into \mathbb{C}

And there is an inverse function , g, that maps a complex domain D into U (but not necessarily onto).

Hence f(g(z)) = z, for all of the z in the domain, D, of g.

Then g is analytic in its domain D, and the following property holds

g '(z) = \frac{1}{f'(g(z))} for z \in D

<< Application to f(z) = z^2 >>

 

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