Intro2ComplexAnalysis

# D since u x 3 v 1 y 3 u x 3 x 2 u y 0 v x 0 v y 31 y

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(d). Since u = x 3 , v = (1 y ) 3 , u x = 3 x 2 , u y = 0 , v x = 0 , v y = 3(1 y ) 2 the function is di ff erentiable only when 3 x 2 = 3(1 y ) 2 or x = 0 , y = 1 , and f ( z ) = 0 . 7.4. (a). f is di ff erentiable at every point on the x and y axes, f ( x + i 0) = 3 x 2 3 , f (0 + iy ) = 3 y 2 3 , not analytic anywhere. (b). f is di ff erentiable only at the point z = 3 16 1 16 i, f ( 3 16 1 16 i ) = 1 4 3 4 i, not analytic anywhere. (c). f is di ff erentiable everywhere, f ( z ) = (6 x + 2) + i (6 y ) , analytic everywhere (entire). (d). f is di ff erentiable for all z ̸ = 0 , ± 2 i,

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50 Lecture 7 f ( z ) = 2( z 4 + 5 z 2 + 12) / [ z 2 ( z 2 + 4) 2 ] , analytic in C { 0 , ± 2 i } . (e). f is di ff erentiable everywhere, f ( z ) = 2 ze z 2 , analytic everywhere (entire). 7.5. a = 1 , b = 3 , c = 3 , y = 0 gives f ( x ) = ix 3 , and hence f ( z ) = iz 3 and dw/dz = 3 iz 2 . 7.6. u x = 2 x, u y = 2 y v y = 2 x and v x = 2 y v = 2 xy + f ( y ) and v = 2 xy + g ( x ) , which is impossible. 7.7. LHS = 2( u 2 x + u 2 y ) + 2( v 2 x + v 2 y ) . 7.8. (a). u ( x, y ) = ( x 3 3 xy 2 ) / ( x 2 + y 2 ) , ( x, y ) ̸ = (0 , 0) 0 , ( x, y ) = (0 , 0) and v ( x, y ) = ( y 3 3 x 2 y ) / ( x 2 + y 2 ) , ( x, y ) ̸ = (0 , 0) 0 , ( x, y ) = (0 , 0) . u x (0 , 0) = 1 , u y (0 , 0) = 0 , v x (0 , 0) = 0 , v y (0 , 0) = 1 . (b). For z ̸ = 0 , ( f ( z ) f (0)) / ( z 0) = ( z/z ) 2 . Now see Problem 5.6 (a). 7.9. Use rules for di ff erentiation. 7.10. Since u r = u x x r + u y y r , u θ = u x x θ + u y y θ , we have u r = u x cos θ + u y sin θ , u θ = u x r sin θ + u y r cos θ (7.9) and v r = v x cos θ + v y sin θ , v θ = v x r sin θ + v y r cos θ , which in view of (6.5) is the same as v r = u y cos θ + u x sin θ , v θ = u y r sin θ + u x r cos θ . (7.10) From (7.9) and (7.10), the Cauchy-Riemann conditions (7.7) are immediate. Now, since f ( z ) = u x + iv x and u x = u r cos θ u θ sin θ r = u r cos θ + v r sin θ , v x = v r cos θ v θ sin θ r = v r cos θ u r sin θ , it follows that f ( z ) = u r (cos θ i sin θ ) + iv r (cos θ i sin θ ) = e i θ ( u r + iv r ) . Since u = r cos θ 2 , v = r sin θ 2 , u r = 1 2 r cos θ 2 , u θ = 1 2 r sin θ 2 , v r = 1 2 r sin θ 2 , v θ = 1 2 r cos θ 2 , and hence conditions (7.7) are satisfied. Thus, f is di ff erentiable at all z except z = 0 . Furthermore, from (7.8) it follows that f ( z ) = e i θ 1 2 r (cos θ 2 + i sin θ 2 ) = 1 2 z . 7.11. Since x = ( z + z ) / 2 and y = ( z z ) / 2 i, u and v may be regarded as functions of z and z. Thus, condition w z = 0 is the same as u x x z + u y y z + i v x x z + v y y z = 0 , which is the same as 1 2 u x 1 2 i u y + i 2 v x 1 2 v y = 0 . Now compare the real and imaginary parts. 7.12. Let f = u + iv. From calculus, it su ces to show that Jacobian J ( x 0 , y 0 ) = u x v x u y v y ( x 0 , y 0 ) ̸ = 0 . Now use (6.5) and (6.3) to obtain J ( x 0 , y 0 ) = u 2 x ( x 0 , y 0 ) + v 2 x ( x 0 , y 0 ) = | f ( z 0 ) | 2 ̸ = 0 . 7.13. (a). i/ 2 , (b). 18 , (c). 16 . 7.14. f = u + iv, u = c u x = u y = 0 v x = v y = 0 v is also a constant, and hence f is a constant. 7.15. If f = u + iv, then u = 2 h 3 and v = h. Now, by the Cauchy-Riemann conditions, we have 6 h 2 h x = h y and 6 h 2 h y = h x . Thus, 12 h 4 h x = h x , or h x (12 h 4 + 1) = 0 , which implies that h x = 0 , and from this h y = 0 .
Analytic Functions II 51 7.16. In (6.1), if we allow z 0 along the x -axis, then f ( z 0 ) is real.

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