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

[jaus tail]

Homework Statement

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 ?

Homework Equations

Z = x + i y = [ sqrt {x² + y² } ] e^{i (theta)}
where theta is tan^-1 (y/x)

The Attempt at a Solution

Function w is
w = ln (z)
= ln( sqrt {[x² + y² ] } multiply with e^{i (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 e^c.

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.

But there is log in u.

 

[pasmith]

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

 

[jaus tail]

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

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

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

Like left hand side is u and not e^u.

 

[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 ) = e^c
so squaring we get
x² + y² = e^c(2) --- equation of circle. radius e^c
that is x² = e^2c - y²

Thanks.

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