# Function in W plane gets mapped to the Z plane as?

1. Feb 5, 2017

### jaus tail

1. The problem statement, all variables and given/known data
A function is given as W = ln (Z)
Z = x + i y
W = u + i v
i is sqrt of (-1)
The u = constant lines get mapped in z plane as ?

2. Relevant equations
Z = x + i y = [ sqrt {x2 + y2 } ] ei (theta)
where theta is tan-1 (y/x)

3. The attempt at a solution
Function w is
w = ln (z)
= ln( sqrt {[x2 + y2 ] } multiply with ei (theta)

Now ln (a b) = ln (a) + ln (b)

so w = ln (sqrt [x2 + y2 ]) + i(theta) ln (e)
The first part is real, while second part is imaginary.
w = u + iv.
So u = ln(sqrt [x2 + y2 ] )
Question says u = constant lines get mapped in z plane as ?
u = constant
so ln (sqrt (x2 + y2 ) = constant.

Now how to proceed? In answer they've said concentric circles with radius ec.

I got till u = ln (sqrt (x2 + y2)
remove sqrt out as 1/2 we get:
u = 1/2 ( ln (x2 + y2) )
equation of circle is x2 + y2 = constant.

2. Feb 5, 2017

### pasmith

Do you not recognize $\ln\sqrt{x^2 + y^2} =\ln |z| = c$ as $|z| = e^c$, which is a circle of radius $e^c$?

3. Feb 5, 2017

### jaus tail

But question asks for mapping of lines u = constant.
And not mapping of lines eu = constant.

I understand how you removed log part by using exponential, but we need u = constant,
and not eu = e constant

Like left hand side is u and not eu.

4. Feb 5, 2017

### jaus tail

Ok now i get it. Mapping on z plane, means we need 'x' in terms of 'y'
Equation that we get is: ln sqrt (x2 + y2 ) = constant c
to get x in terms of y,
sqrt (x2 + y2 ) = ec
so squaring we get
x2 + y2 = ec(2) --- equation of circle. radius ec
that is x2 = e2c - y2

Thanks.

PS how do you put the sign of square root over x2 + y2