HW4sol - Log z 3 = ln | z 3 | +Arg z 3 is not continuous...

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P ROBLEM SET 4 Partial solution 1. By Cauchy–Riemann equations (since u and v have continuous partial derivatives) for f to be analytic it suffices that for all u x = 3 x 2 + ay 2 = bx 2 + 3 cy 2 = v y and u y = 2 axy = - 2 bxy = - v x . Therefore 3 = b , a = 3 c and a = - b . So a = - 3 , b = 3 , c = - 1 , and f 0 ( z ) = u x + iv x = 3 x 2 - 3 y 3 + 6 xyi . 2. (Sec 2.1 problem 15). Let f = u + iv and f 0 = 0 in D . That is f 0 = u x - iu y = 0 . Hence u x = u y = 0 . We know from multivariable calculus that a function with real values (like u ) that has partial derivatives equal to 0 on a connected open set (like D ) must be a constant. Therefore u is a constant. Since f 0 can also be writen as f 0 = v y + iv x it follows that v x = v y = 0 . Therefore, as above that v is a constant. Hence f = u + iv is a constant. 5. Since h is harmonic h xx = - h yy . It suffices to check that the Cauchy-Riamann equations are satisfied. f = u + iv = h x + i ( - h y ) . Note that u x = h xx = - h yy = v y and u y = h xy = - v x . 6. From the definition of the argument follows that Arg z 3 = 3 Arg z + 2 π if Arg z [ - π, - π/ 3) 3 Arg z if Arg z [ - π/ 3 , π/ 3) 3 Arg z - 2 π if Arg z [ π/ 3 , π ) Therefore
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Unformatted text preview: Log z 3 = ln | z 3 | +Arg z 3 is not continuous when the argument of z is equal to- ,-/ 3 or / 3 , or if z = 0 . At all other points it is continuous and moreover analytic since it is as a composition of two analytic functions. To be more precise recall that by the chain rule [ f ( g ( z ))] = f ( g ( z )) g ( z ) . In our case f ( w ) = Log w and g ( z ) = z 3 . Recall that derivarive of Log w exists when Arg w 6 =- and w 6 = 0 . Hence the f ( z 3 ) exists if Arg z 3 6 =- and z 6 = 0 . From the above we see that Arg z 3 =- if Arg z =- , / 3 or / 3 . So Log z 3 is analitic on the set of complex numbers that are different from zero and whose argument is not equal to- ,-/ 3 , or / 3 ....
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