Complex Function Derivatives and Analytic Functions, Lecture notes of Engineering Science and Technology

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MAT 247 Engineering Mathematics

Hakkı Ula¸s Unal¨ Dept. of Electrical-Electronics Eng. Eski¸sehir Technical University, Turkey

November 12, 2018

Today

Derivatives

I (^) Let f (z) be a function, whose domain of definition contains a point in a neighborhood of zo. Then, derivative of f (z), which is shown as f ′(zo), is defined as

f ′(zo) := lim z→z o

f (z) − f (zo) z − zo

I (^) f is called differentiable at zo if f ′(zo) exists.

I (^) Let f (z) be a function, whose domain of definition contains a point in a neighborhood of zo. Then, derivative of f (z), which is shown as f ′(zo), is defined as

f ′(zo) := lim z→z o

f (z) − f (zo) z − zo

I (^) f is called differentiable at zo if f ′(zo) exists.

I (^) Let ∆z := z − zo, where z 6 = zo. Then,

f ′(zo) := (^) ∆limz→ 0

f (zo + ∆z) − f (zo) ∆z.

I (^) Let f (z) be a function, whose domain of definition contains a point in a neighborhood of zo. Then, derivative of f (z), which is shown as f ′(zo), is defined as

f ′(zo) := lim z→z o

f (z) − f (zo) z − zo

I (^) f is called differentiable at zo if f ′(zo) exists.

I (^) Let ∆z := z − zo, where z 6 = zo. Then,

f ′(zo) := (^) ∆limz→ 0

f (zo + ∆z) − f (zo) ∆z.

I (^) f (zo + ∆z) is it defined? (Note:f is defined in some neighbourhood of zo ) I (^) Let define ∆w := f (zo + ∆z) − f (zo), then, f ′(z) at z = zo is

dw dz

= (^) ∆limz→ 0 ∆w ∆z

Examples

Ex

Find the derivative of f (z) = z, if exists. I ∆w ∆z =

(z + ∆z) − z ∆z =

∆z ∆z ,

Examples

Ex

Find the derivative of f (z) = z, if exists. I ∆w ∆z =

(z + ∆z) − z ∆z =

∆z ∆z , I (^) Note, dwdz = lim∆z→ 0 ∆ ∆wz , where

∆z → 0 ⇒ (∆x, ∆y) → 0 ,

I (^) Let (∆x, ∆y) → 0 along the real axis, that is,

∆z → 0 ⇒ (∆x, 0) → 0 ,

Examples

Ex

Find the derivative of f (z) = z, if exists. I ∆w ∆z =

(z + ∆z) − z ∆z =

∆z ∆z , I (^) Note, dwdz = lim∆z→ 0 ∆ ∆wz , where

∆z → 0 ⇒ (∆x, ∆y) → 0 ,

I (^) Let (∆x, ∆y) → 0 along the real axis, that is,

∆z → 0 ⇒ (∆x, 0) → 0 ,

I (^) ∆z = ∆x + j0 = ∆x + j0 = ∆z, hence, ∆ ∆wz = ∆ ∆zz = 1,

Examples

Ex

Find the derivative of f (z) = z, if exists. I ∆w ∆z =

(z + ∆z) − z ∆z =

∆z ∆z , I (^) Note, dwdz = lim∆z→ 0 ∆ ∆wz , where

∆z → 0 ⇒ (∆x, ∆y) → 0 ,

I (^) Let (∆x, ∆y) → 0 along the real axis, that is,

∆z → 0 ⇒ (∆x, 0) → 0 ,

I (^) ∆z = ∆x + j0 = ∆x + j0 = ∆z, hence, ∆ ∆wz = ∆ ∆zz = 1, I (^) Similarly, ∆z = 0 + j∆y = 0 − j∆y = −∆z, hence, ∆ ∆wz = ∆ ∆zz = − 1 I (^) Remember the uniqueness of limit.

Example

Ex-

Show that f (z) = |z|^2 is only differentiable at point z = 0.

Differentiation rules

I (^) Let f be a differentiable function at a point z and c be a complex constant d dz

c = 0 d dz

z = 1 d dz

(cf (z)) = c d dz

f (z) d dz

zn^ = nz(n−1) I (^) Let f and g be differentiable functions at a point z, then, d dz

(f (z) + g(z)) = d dz

f (z) + d dz

g(z),

d dz (f^ (z)g(z)) =^ f^ (z)^

d dz g(z) +^ g(z)^

d dz f^ (z) d dz

f (z) g(z)

g(z) (^) dzd f (z) − f (z) (^) dzd g(z) g^2 (z)

Differentiation rules

I (^) Let f be a differentiable function at a point z and c be a complex constant d dz

c = 0 d dz

z = 1 d dz

(cf (z)) = c d dz

f (z) d dz

zn^ = nz(n−1) I (^) Let f and g be differentiable functions at a point z, then, d dz

(f (z) + g(z)) = d dz

f (z) + d dz

g(z),

d dz (f^ (z)g(z)) =^ f^ (z)^

d dz g(z) +^ g(z)^

d dz f^ (z) d dz

f (z) g(z)

g(z) (^) dzd f (z) − f (z) (^) dzd g(z) g^2 (z)

I (^) Suppose that f ′(zo) exists and g can be differentiable at f (zo). Then, F (z) = g(f (z)) can be differentiable at zo and it equals to F ′(zo) = f ′(zo)g′(f (zo))

Cauchy-Reimann Equations

I (^) Let f (z) = u(x, y) + jv(x, y) be a differentiable function at zo, where zo = xo + jyo. Then, since

f ′(zo) := lim z→zo^ f^ (z z) −−^ fz^ (zo) o

exists, by letting ∆z := z − zo =: ∆x + j∆y ∆w := f (zo + ∆z) − f (zo),

f ′(zo) = (^) ∆limz→ 0

∆w ∆z.

Cauchy-Reimann Equations

I (^) Let f (z) = u(x, y) + jv(x, y) be a differentiable function at zo, where zo = xo + jyo. Then, since

f ′(zo) := lim z→zo^ f^ (z z) −−^ fz^ (zo) o

exists, by letting ∆z := z − zo =: ∆x + j∆y ∆w := f (zo + ∆z) − f (zo),

f ′(zo) = (^) ∆limz→ 0

∆w ∆z. I (^) Note, ∆w =f (zo + ∆z) − f (zo) (1) =u(xo + ∆x, yo + ∆y) − u(xo, yo)

• j (v(xo + ∆x, yo + ∆y) − v(xo, yo)) , (2) limz→zo f (z) = wo iff lim (x,y)→(xo,yo)

u(x, y) = uo lim (x,y)→(xo,yo)

v(x, y) = vo.