Q6P - Mathematical Methods in Physical Sciences

Question

w = √z. Hint: This is equivalent to w² = z; find x and y in terms of u and v and then solve the pair of equations for u and v in terms of x and y. Note that this is really the same problem as Problem 1 with the z and w planes interchanged.

Step-by-Step Solution

Answer

The answer is 

$u^2=\frac{x+\sqrt{x^2+y^2}}{2}$ and  $v^2=\frac{-x+\sqrt{x^2+y^2}}{2}$.

Step 1: Function is given as w = √ z

Function can be written as:

         $w^2=z$                                                                                                                ...... (1)

From Cauchy Riemann hypothesis:

        $w=u+iv$                                                                                                           ...... (2)

          $z=x+iy$                                                                                                          ...... (3)

 

Put both (2) and (3) value in equation (1) as follows:

 $$\begin{gathered} x+iy={\left(u+iv\right)}^2 \\ x+iy=u^2-v^2+2uvi \end{gathered}$$

Step 2: Compare the real and imaginary term

Comparing real and imaginary term as follows:

  $u^2-v^2=x$                                                                                                               ...... (4)

      $y=2uv$                                                                                                                ...... (5)

Equation (5) can be expressed as follows:

       $u=\frac{y}{2v}$                                                                                                                 ...... (6)

     $v=\frac{y}{2u}$                                                                                                                   ...... (7)

 

Put (7) in equation (4) as follows:

   $u^2-{\left(\frac{y}{2u}\right)}^2=x$                                                                                                       ...... (8)

Put (6) in equation (4) as follows:

      ${\left(\frac{y}{2v}\right)}^2-v^2=x$                                                                                                    ...... (9)

Solve equation (8) as follows:

 $\begin{array}{rcl}{u}^2-{\left(\frac{y}{2u}\right)}^2& =& x\\ 4v^2-y^2& =& 4x{v}^2\\ 4v^4-4x{v}^2-y^2& =& 0\end{array}$

Here,  

a = 4, b = -4x , c = -y²

 

 $\begin{array}{rcl}{v}^2& =& \frac{x+\sqrt{x^2+y^2}}{2}\end{array}$

Here rejecting the negative solution since value of the u should be real.

Now solving another equation no (9) as follows:

 $\begin{array}{rcl}{\left(\frac{y}{2v}\right)}^2-v^2& =& x\\ 4v^4-y^2& =& -4x{v}^2\\ 4v^4+4x{v}^2-y^2& =& 0\end{array}$

Here, 

a = 4, b = 4x, c = - y² .

Solution of the equation will be as follows:

 $\begin{array}{rcl}{v}^2& =& \frac{-x\sqrt{x^2+y^2}}{2}\end{array}$

Here rejecting the negative solution since value of the v should be real.

Hence,  

$\begin{array}{rcl}{u}^2& =& \frac{x\sqrt{x^2+y^2}}{2},v^2=\frac{-x\sqrt{x^2+y^2}}{2}\end{array}$and .