f(x + h) f(x) h as representing the slope of a secant line. As h goes to 0, the slope of the secant line approaches the slope of the tangent line.

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1 Derivative of f(z) Dr. E. Jacobs Te erivative of a function is efine as a limit: f (x) 0 f(x + ) f(x) We can visualize te expression f(x+) f(x) as representing te slope of a secant line. As goes to 0, te slope of te secant line approaces te slope of te tangent line. You learne in Calculus I tat te limit efinition is sufficient to calculate erivatives of elementary functions. For example, to fin te erivative of f(x) = x 3, we perform te following limit calculation: f (x) 0 (x + ) 3 x 3 0 x 3 + 3x 2 + 3x x 3 0 ( 3x 2 + 3x + 2) = 3x 2 We may apply exactly te same limit process to fin te erivative of a complex value function. If f(z) is a function of a complex variable ten we efine: f f(z + z) f(z) (z)

2 Since f(z) is actually a vector fiel, tere is no tangent line interpretation of te erivative ere. However, te limit calculations still work te same way. For example, suppose f(z) = z 3. Fin f (z) f (z + z) 3 z 3 (z) z 3 + 3z 2 z + 3z( z) 2 + ( z) 3 z 3 z 0 z ( 3z 2 + 3z z + ( z) 2) z 0 = 3z 2 All te familiar erivative formulas are consequences of te limit efinition. Consequently, we can sow eac of te following familiar erivative formulas using limits: z (zn ) = nz n 1 (sin z) = cos z z (cos z) = sin z z (sin z) = cos z z (cos z) = sin z z z (ez ) = e z z (f(z) + g(z)) = f (z) + g (z) z (f(z) g(z)) = f(z)g (z) + f (z)g(z) ( ) f(z) = g(z)f (z) f(z)g (z) z g(z) (g(z)) 2 z (f(g(z)) = f (g(z)) g (z)

3 Wen we take 0 to fin te erivative, it is not suppose to matter weter is positive or negative. f(x + ) f(x) lim f(x + ) f(x) For a function of a complex variable f(z), wen we fin te erivative by letting z 0, te irection of z is not suppose to matter. However, since z is a vector in two imensions, tere are now many more irections tat z can approac 0 from. Te conition tat we are suppose to get te same answer no matter wat irection z is pointing leas to an important pair of equations calle te Caucy- Riemann equations. Let us begin by writing bot z an f(z) in terms of teir real an imaginary parts. z = x + iy f(z) = u(x, y) + iv(x, y) Bot z an f(z + z) f(z) will ave real an imaginary parts. z = x + i y f(z + z) f(z) = u + i v

4 Te erivative can now be written in te following form: f f(z + z) f(z) (z) z 0 u + i v z Let s take te case were z approaces 0 in a irection parallel to te real axis. z = x + i y = x + i 0 = x f u + i v (z) x 0 ( u x + i v x ) x 0 u + i v x = Now, let s take te case were z approaces 0 in a irection parallel to te imaginary axis. Of course 1 i = 1 i i i = i i 2 z = x + i y = 0 + i y = i y f u + i v (z) y 0 ( 1 i = i so, f (z) = 1 i u y + v y u + i v y 0 i y ) = 1 i + + = i + Tis leas to two ifferent formulas for te erivative tat are suppose to give te same answer. f (z) = f (z) = i As an example, consier te function f(z) = z 3 wose erivative f (z) = 3z 2 was calculate earlier using limits. Tis time, let s try oing te calculation using f (z) =.

5 Since z 3 = (x + iy) 3 = x 3 3xy 2 + i(3x 2 y y 3 ) it follows tat: u = x 3 3xy 2 v = 3x 2 y y 3 = 3x2 3y 2 Terefore, te erivative must be: = 6xy f (z) = = 3(x 2 y 2 ) + 6xyi = 3 ( x 2 + 2xyi y 2) = 3(x + iy) 2 = 3z 2 Tis is te same formula obtaine earlier. We coul ave just as easily use te formula f (z) = i f (z) = i = ( 3x 2 y y 3) i ( x 3 3xy 2) = 3x 2 3y 2 i ( 6xy) = 3 ( x 2 + 2xy y 2) = 3(x + iy) 2 = 3z 2 Since = i ten we may compare te real parts of bot sies of te equation an te imaginary parts of bot sies to obtain: = = Tese are calle te Caucy-Riemann Equations an tey are a conition for ifferentiability. If te Caucy-Riemann Equations are not true for a function f(z) ten it s erivative oesn t exist. Example. Consier te function f(z) = z 2. In tis case, u(x, y) = x 2 + y 2 an v(x, y) = 0. = 2x will only equal = 0 at x = 0. = 0 will only equal = 2y at y = 0. Consequently, te function f(z) = z 2 as a erivative at z = 0 but nowere else.

6 Example. Does f(z) = e z ave a erivative? If so, wat is it? We first ientify u an v an ten see if te Caucy-Riemann equations are satisfie. e z = e x+iy = e x e iy = e x (cos y + i sin y) = e x cos y + ie x sin y u = e x cos y v = e x sin y = ex sin y is te same as at all points. Also, = ex sin y is te same as = ( ex sin y) = e x sin y at all points. Terefore, te erivative exists at all points an is given by: f (z) = = ex cos y + ie x sin y = e x (cos y + i sin y) = e x+iy = e z Definition A function f(z) is sai to be analytic at a point z 0 if te erivative of f(z) exists at z 0 as well as in some isk aroun z 0. For example, te erivative of z 2 exists at 0 but nowere else, so z 2 is not analytic at 0. On te oter an, te erivative of e z exists not only at 0 but in all points in any isk surrouning 0. Terefore e z is analytic at 0. Of course, in te case of e z, te function is analytic at all oter points as well. In te next sections, you ll see ow te teory of analytic functions an te Caucy-Riemann equations ave applications to calculations involving flui flow.