Express z = − 3 + i z = -\sqrt{3} + i z = − 3 + i z 1 / 3 z^{1/3} z 1/3
Solution ∣ z ∣ = 3 + 1 = 2 |z| = \sqrt{3 + 1} = 2 ∣ z ∣ = 3 + 1 = 2 R e ( z ) < 0 \mathrm{Re}(z) \lt 0 Re ( z ) < 0 I m ( z ) > 0 \mathrm{Im}(z) \gt 0 Im ( z ) > 0 arg ( z ) = π − π / 6 = 5 π / 6 \arg(z) = \pi - \pi/6 = 5\pi/6 arg ( z ) = π − π /6 = 5 π /6
z = 2 e 5 π i / 6 z = 2\,e^{5\pi i/6} z = 2 e 5 π i /6
z 1 / 3 = 2 1 / 3 e ( 5 π / 6 + 2 π k ) / 3 z^{1/3} = 2^{1/3}\, e^{(5\pi/6 + 2\pi k)/3} z 1/3 = 2 1/3 e ( 5 π /6 + 2 π k ) /3 k = 0 , 1 , 2 k = 0, 1, 2 k = 0 , 1 , 2
z 0 = 2 1 / 3 e 5 π i / 18 z_0 = 2^{1/3}\, e^{5\pi i/18} z 0 = 2 1/3 e 5 π i /18 z 1 = 2 1 / 3 e 17 π i / 18 z_1 = 2^{1/3}\, e^{17\pi i/18} z 1 = 2 1/3 e 17 π i /18 z 2 = 2 1 / 3 e 29 π i / 18 z_2 = 2^{1/3}\, e^{29\pi i/18} z 2 = 2 1/3 e 29 π i /18
If you get this wrong, revise: Section 1.5 (Roots of Complex Numbers).
Let f ( z ) = z 2 + z ˉ 2 f(z) = z^2 + \bar{z}^2 f ( z ) = z 2 + z ˉ 2 f f f
Solution f ( z ) = ( x + i y ) 2 + ( x − i y ) 2 = 2 ( x 2 − y 2 ) f(z) = (x + iy)^2 + (x - iy)^2 = 2(x^2 - y^2) f ( z ) = ( x + i y ) 2 + ( x − i y ) 2 = 2 ( x 2 − y 2 ) u = 2 ( x 2 − y 2 ) u = 2(x^2 - y^2) u = 2 ( x 2 − y 2 ) v = 0 v = 0 v = 0
u x = 4 x u_x = 4x u x = 4 x u y = − 4 y u_y = -4y u y = − 4 y v x = 0 v_x = 0 v x = 0 v y = 0 v_y = 0 v y = 0
CR: 4 x = 0 ⇒ x = 0 4x = 0 \Rightarrow x = 0 4 x = 0 ⇒ x = 0 − 4 y = 0 ⇒ y = 0 -4y = 0 \Rightarrow y = 0 − 4 y = 0 ⇒ y = 0
f f f z = 0 z = 0 z = 0
f " ( 0 ) = 0 f"(0) = 0 f " ( 0 ) = 0
If you get this wrong, revise: Sections 2.4 and 3.1 (Analyticity and Cauchy-Riemann).
Verify that f ( z ) = 1 z 2 + 1 f(z) = \frac{1}{z^2 + 1} f ( z ) = z 2 + 1 1 f ′ ( z ) f'(z) f ′ ( z )
Solution f ( z ) = 1 / ( z 2 + 1 ) f(z) = 1/(z^2 + 1) f ( z ) = 1/ ( z 2 + 1 ) ± i \pm i ± i f f f C ∖ { i , − i } \mathbb{C} \setminus \{i, -i\} C ∖ { i , − i }
By the quotient rule: f ′ ( z ) = − 2 z ( z 2 + 1 ) 2 f'(z) = \frac{-2z}{(z^2 + 1)^2} f ′ ( z ) = ( z 2 + 1 ) 2 − 2 z
Verify via CR at z = 1 z = 1 z = 1 u = x 2 − y 2 + 1 ( x 2 − y 2 + 1 ) 2 + 4 x 2 y 2 u = \frac{x^2 - y^2 + 1}{(x^2 - y^2 + 1)^2 + 4x^2y^2} u = ( x 2 − y 2 + 1 ) 2 + 4 x 2 y 2 x 2 − y 2 + 1 v = − 2 x y ( x 2 − y 2 + 1 ) 2 + 4 x 2 y 2 v = \frac{-2xy}{(x^2 - y^2 + 1)^2 + 4x^2y^2} v = ( x 2 − y 2 + 1 ) 2 + 4 x 2 y 2 − 2 x y
u x ( 1 , 0 ) = − 1 / 2 = f ′ ( 1 ) u_x(1, 0) = -1/2 = f'(1) u x ( 1 , 0 ) = − 1/2 = f ′ ( 1 ) ✓ \checkmark ✓
If you get this wrong, revise: Sections 3.1 and 3.3 (CR Equations).
Show that u ( x , y ) = x 3 − 3 x y 2 + 3 x 2 − 3 y 2 u(x, y) = x^3 - 3xy^2 + 3x^2 - 3y^2 u ( x , y ) = x 3 − 3 x y 2 + 3 x 2 − 3 y 2
Solution u x x = 6 x + 6 u_{xx} = 6x + 6 u xx = 6 x + 6 u y y = − 6 x − 6 u_{yy} = -6x - 6 u y y = − 6 x − 6 Δ u = 0 \Delta u = 0 Δ u = 0 ✓ \checkmark ✓
By CR: v y = u x = 3 x 2 − 3 y 2 + 6 x v_y = u_x = 3x^2 - 3y^2 + 6x v y = u x = 3 x 2 − 3 y 2 + 6 x v = 3 x 2 y − y 3 + 6 x y + g ( x ) v = 3x^2 y - y^3 + 6xy + g(x) v = 3 x 2 y − y 3 + 6 x y + g ( x )
v x = − u y = 6 x y + 6 y v_x = -u_y = 6xy + 6y v x = − u y = 6 x y + 6 y 6 x y + 6 y = 6 x y + 6 y + g ′ ( x ) ⇒ g ′ ( x ) = 0 ⇒ g ( x ) = C 6xy + 6y = 6xy + 6y + g'(x) \Rightarrow g'(x) = 0 \Rightarrow g(x) = C 6 x y + 6 y = 6 x y + 6 y + g ′ ( x ) ⇒ g ′ ( x ) = 0 ⇒ g ( x ) = C
Harmonic conjugate: v ( x , y ) = 3 x 2 y − y 3 + 6 x y + C v(x, y) = 3x^2 y - y^3 + 6xy + C v ( x , y ) = 3 x 2 y − y 3 + 6 x y + C
f ( z ) = u + i v = z 3 + 3 z 2 f(z) = u + iv = z^3 + 3z^2 f ( z ) = u + i v = z 3 + 3 z 2
If you get this wrong, revise: Section 3.4 (Harmonic Functions).
Evaluate ∫ γ ( z 2 + 2 z ) d z \int_\gamma (z^2 + 2z)\, dz ∫ γ ( z 2 + 2 z ) d z γ \gamma γ z = 1 z = 1 z = 1 z = − 1 z = -1 z = − 1
Solution Since z 2 + 2 z z^2 + 2z z 2 + 2 z F ( z ) = z 3 / 3 + z 2 F(z) = z^3/3 + z^2 F ( z ) = z 3 /3 + z 2
∫ γ ( z 2 + 2 z ) d z = F ( − 1 ) − F ( 1 ) = 2 3 − 4 3 = − 2 3 \int_\gamma (z^2 + 2z)\, dz = F(-1) - F(1) = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3} ∫ γ ( z 2 + 2 z ) d z = F ( − 1 ) − F ( 1 ) = 3 2 − 3 4 = − 3 2
If you get this wrong, revise: Sections 4.5 and 4.7 (Contour Integrals).
Use the ML inequality to bound ∣ ∫ γ e z z − 2 d z ∣ \left|\int_\gamma \frac{e^z}{z - 2}\, dz\right| ∫ γ z − 2 e z d z γ \gamma γ ∣ z ∣ = 1 |z| = 1 ∣ z ∣ = 1
Solution On γ \gamma γ ∣ z ∣ = 1 |z| = 1 ∣ z ∣ = 1 ∣ e z ∣ ≤ e |e^z| \leq e ∣ e z ∣ ≤ e ∣ z − 2 ∣ ≥ 1 |z - 2| \geq 1 ∣ z − 2∣ ≥ 1
∣ e z z − 2 ∣ ≤ e \left|\frac{e^z}{z - 2}\right| \leq e z − 2 e z ≤ e L = 2 π L = 2\pi L = 2 π
∣ ∫ γ e z z − 2 d z ∣ ≤ 2 π e \left|\int_\gamma \frac{e^z}{z - 2}\, dz\right| \leq 2\pi e ∫ γ z − 2 e z d z ≤ 2 π e
If you get this wrong, revise: Section 4.6 (ML Inequality).
Evaluate ∮ γ z + 1 z 2 − z d z \oint_\gamma \frac{z + 1}{z^2 - z}\, dz ∮ γ z 2 − z z + 1 d z γ \gamma γ ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Solution z + 1 z 2 − z = z + 1 z ( z − 1 ) \frac{z + 1}{z^2 - z} = \frac{z + 1}{z(z - 1)} z 2 − z z + 1 = z ( z − 1 ) z + 1 z = 0 z = 0 z = 0 z = 1 z = 1 z = 1 ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
At z = 0 z = 0 z = 0 R e s = lim z → 0 z + 1 z − 1 = − 1 \mathrm{Res} = \lim_{z \to 0} \frac{z + 1}{z - 1} = -1 Res = lim z → 0 z − 1 z + 1 = − 1 z = 1 z = 1 z = 1 R e s = lim z → 1 z + 1 z = 2 \mathrm{Res} = \lim_{z \to 1} \frac{z + 1}{z} = 2 Res = lim z → 1 z z + 1 = 2
∮ γ z + 1 z 2 − z d z = 2 π i ( − 1 + 2 ) = 2 π i \oint_\gamma \frac{z + 1}{z^2 - z}\, dz = 2\pi i(-1 + 2) = 2\pi i ∮ γ z 2 − z z + 1 d z = 2 π i ( − 1 + 2 ) = 2 π i
If you get this wrong, revise: Sections 8.4 and 8.5 (Residues).
Classify all singularities of f ( z ) = e 1 / z z 2 + 1 f(z) = \frac{e^{1/z}}{z^2 + 1} f ( z ) = z 2 + 1 e 1/ z
Solution z = 0 z = 0 z = 0 e 1 / z e^{1/z} e 1/ z 0 0 0 z = 0 z = 0 z = 0 f f f z = i z = i z = i z = − i z = -i z = − i
At z = i z = i z = i R e s = e 1 / i 2 i = e − i 2 i \mathrm{Res} = \frac{e^{1/i}}{2i} = \frac{e^{-i}}{2i} Res = 2 i e 1/ i = 2 i e − i z = − i z = -i z = − i R e s = e 1 / ( − i ) − 2 i = e i − 2 i \mathrm{Res} = \frac{e^{1/(-i)}}{-2i} = \frac{e^{i}}{-2i} Res = − 2 i e 1/ ( − i ) = − 2 i e i
At z = 0 z = 0 z = 0 1 / z 1/z 1/ z e 1 / z z 2 + 1 \frac{e^{1/z}}{z^2 + 1} z 2 + 1 e 1/ z 1 z 2 + 1 = 1 − z 2 + z 4 − ⋯ \frac{1}{z^2 + 1} = 1 - z^2 + z^4 - \cdots z 2 + 1 1 = 1 − z 2 + z 4 − ⋯ z = 0 z = 0 z = 0 e 1 / z = 1 + 1 / z + 1 / ( 2 z 2 ) + ⋯ e^{1/z} = 1 + 1/z + 1/(2z^2) + \cdots e 1/ z = 1 + 1/ z + 1/ ( 2 z 2 ) + ⋯ 1 / z 1/z 1/ z 1 ⋅ 1 / z = 1 / z 1 \cdot 1/z = 1/z 1 ⋅ 1/ z = 1/ z 1 1 1
If you get this wrong, revise: Sections 8.1 and 8.4 (Singularities and Residues).
Evaluate ∫ 0 2 π cos θ 5 + 4 cos θ d θ \int_0^{2\pi} \frac{\cos\theta}{5 + 4\cos\theta}\, d\theta ∫ 0 2 π 5 + 4 c o s θ c o s θ d θ
Solution Substitute z = e i θ z = e^{i\theta} z = e i θ
I = ∫ ∣ z ∣ = 1 ( z + z − 1 ) / 2 5 + 4 ( z + z − 1 ) / 2 ⋅ d z i z = 1 2 i ∫ ∣ z ∣ = 1 z 2 + 1 z ( 2 z 2 + 5 z + 2 ) d z = 1 2 i ∫ ∣ z ∣ = 1 z 2 + 1 z ( 2 z + 1 ) ( z + 2 ) d z I = \int_{|z|=1} \frac{(z + z^{-1})/2}{5 + 4(z + z^{-1})/2} \cdot \frac{dz}{iz} = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z^2 + 5z + 2)}\, dz = \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 1}{z(2z + 1)(z + 2)}\, dz I = ∫ ∣ z ∣ = 1 5 + 4 ( z + z − 1 ) /2 ( z + z − 1 ) /2 ⋅ i z d z = 2 i 1 ∫ ∣ z ∣ = 1 z ( 2 z 2 + 5 z + 2 ) z 2 + 1 d z = 2 i 1 ∫ ∣ z ∣ = 1 z ( 2 z + 1 ) ( z + 2 ) z 2 + 1 d z
Poles inside ∣ z ∣ = 1 |z| = 1 ∣ z ∣ = 1 z = 0 z = 0 z = 0 z = − 1 / 2 z = -1/2 z = − 1/2
At z = 0 z = 0 z = 0 R e s = 1 ( 2 ⋅ 0 + 1 ) ( 0 + 2 ) = 1 2 \mathrm{Res} = \frac{1}{(2 \cdot 0 + 1)(0 + 2)} = \frac{1}{2} Res = ( 2 ⋅ 0 + 1 ) ( 0 + 2 ) 1 = 2 1 z = − 1 / 2 z = -1/2 z = − 1/2 R e s = 1 / 4 + 1 ( − 1 / 2 ) ( − 1 + 2 ) = 5 / 4 − 1 / 2 = − 5 2 \mathrm{Res} = \frac{1/4 + 1}{(-1/2)(-1 + 2)} = \frac{5/4}{-1/2} = -\frac{5}{2} Res = ( − 1/2 ) ( − 1 + 2 ) 1/4 + 1 = − 1/2 5/4 = − 2 5
I = 1 2 i ⋅ 2 π i ( 1 2 − 5 2 ) = π ( − 2 ) = − π 3 I = \frac{1}{2i} \cdot 2\pi i\left(\frac{1}{2} - \frac{5}{2}\right) = \pi(-2) = -\frac{\pi}{3} I = 2 i 1 ⋅ 2 π i ( 2 1 − 2 5 ) = π ( − 2 ) = − 3 π
If you get this wrong, revise: Section 9.4 (Trigonometric Integrals).
Evaluate ∫ − ∞ ∞ d x ( x 2 + 1 ) ( x 2 + 4 ) \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)(x^2 + 4)} ∫ − ∞ ∞ ( x 2 + 1 ) ( x 2 + 4 ) d x
Solution f ( z ) = 1 ( z 2 + 1 ) ( z 2 + 4 ) f(z) = \frac{1}{(z^2 + 1)(z^2 + 4)} f ( z ) = ( z 2 + 1 ) ( z 2 + 4 ) 1 z = i z = i z = i z = 2 i z = 2i z = 2 i
At z = i z = i z = i R e s = 1 ( 2 i ) ( i 2 + 4 ) = 1 2 i ⋅ 3 = 1 6 i \mathrm{Res} = \frac{1}{(2i)(i^2 + 4)} = \frac{1}{2i \cdot 3} = \frac{1}{6i} Res = ( 2 i ) ( i 2 + 4 ) 1 = 2 i ⋅ 3 1 = 6 i 1 z = 2 i z = 2i z = 2 i R e s = 1 ( 4 i − 1 ) ( 4 i ) = 1 4 i ( − 3 ) = − 1 12 i \mathrm{Res} = \frac{1}{(4i - 1)(4i)} = \frac{1}{4i(-3)} = -\frac{1}{12i} Res = ( 4 i − 1 ) ( 4 i ) 1 = 4 i ( − 3 ) 1 = − 12 i 1
∫ − ∞ ∞ f ( x ) d x = 2 π i ( 1 6 i − 1 12 i ) = 2 π i ⋅ 1 12 i = π 6 \int_{-\infty}^{\infty} f(x)\, dx = 2\pi i\left(\frac{1}{6i} - \frac{1}{12i}\right) = 2\pi i \cdot \frac{1}{12i} = \frac{\pi}{6} ∫ − ∞ ∞ f ( x ) d x = 2 π i ( 6 i 1 − 12 i 1 ) = 2 π i ⋅ 12 i 1 = 6 π
If you get this wrong, revise: Section 9.2 (Rational Function Integrals).
Find the Taylor series of f ( z ) = z z 2 + 4 f(z) = \frac{z}{z^2 + 4} f ( z ) = z 2 + 4 z z 0 = 0 z_0 = 0 z 0 = 0
Solution z z 2 + 4 = z 4 ⋅ 1 1 + z 2 / 4 = z 4 ∑ n = 0 ∞ ( − 1 ) n z 2 n 4 n = ∑ n = 0 ∞ ( − 1 ) n z 2 n + 1 4 n + 1 \frac{z}{z^2 + 4} = \frac{z}{4} \cdot \frac{1}{1 + z^2/4} = \frac{z}{4}\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{4^n} = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{4^{n+1}} z 2 + 4 z = 4 z ⋅ 1 + z 2 /4 1 = 4 z ∑ n = 0 ∞ ( − 1 ) n 4 n z 2 n = ∑ n = 0 ∞ 4 n + 1 ( − 1 ) n z 2 n + 1
For ∣ z ∣ < 2 |z| \lt 2 ∣ z ∣ < 2 0 0 0 ± 2 i \pm 2i ± 2 i 2 2 2
If you get this wrong, revise: Section 7.1 (Taylor Series).
Find the Laurent series of f ( z ) = 1 ( z − 1 ) ( z − 2 ) f(z) = \frac{1}{(z - 1)(z - 2)} f ( z ) = ( z − 1 ) ( z − 2 ) 1 1 < ∣ z ∣ < 2 1 \lt |z| \lt 2 1 < ∣ z ∣ < 2
Solution 1 ( z − 1 ) ( z − 2 ) = 1 z − 2 − 1 z − 1 \frac{1}{(z-1)(z-2)} = \frac{1}{z - 2} - \frac{1}{z - 1} ( z − 1 ) ( z − 2 ) 1 = z − 2 1 − z − 1 1
For ∣ z ∣ > 1 |z| \gt 1 ∣ z ∣ > 1 1 z − 1 = 1 z ⋅ 1 1 − 1 / z = ∑ n = 0 ∞ z − n − 1 \frac{1}{z - 1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=0}^{\infty} z^{-n-1} z − 1 1 = z 1 ⋅ 1 − 1/ z 1 = ∑ n = 0 ∞ z − n − 1
For ∣ z ∣ < 2 |z| \lt 2 ∣ z ∣ < 2 1 z − 2 = − 1 2 ⋅ 1 1 − z / 2 = − ∑ n = 0 ∞ z n 2 n + 1 \frac{1}{z - 2} = -\frac{1}{2} \cdot \frac{1}{1 - z/2} = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} z − 2 1 = − 2 1 ⋅ 1 − z /2 1 = − ∑ n = 0 ∞ 2 n + 1 z n
f ( z ) = − ∑ n = 0 ∞ z n 2 n + 1 − ∑ n = 0 ∞ z − n − 1 f(z) = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} - \sum_{n=0}^{\infty} z^{-n-1} f ( z ) = − ∑ n = 0 ∞ 2 n + 1 z n − ∑ n = 0 ∞ z − n − 1
If you get this wrong, revise: Section 7.4 (Laurent Series).
Using Rouché’s theorem, determine the number of roots of z 5 − 5 z + 1 = 0 z^5 - 5z + 1 = 0 z 5 − 5 z + 1 = 0 ∣ z ∣ < 1 |z| \lt 1 ∣ z ∣ < 1
Solution On ∣ z ∣ = 1 |z| = 1 ∣ z ∣ = 1 ∣ − 5 z ∣ = 5 > ∣ z 5 + 1 ∣ ≤ 2 |-5z| = 5 \gt |z^5 + 1| \leq 2 ∣ − 5 z ∣ = 5 > ∣ z 5 + 1∣ ≤ 2
By Rouché with f ( z ) = − 5 z f(z) = -5z f ( z ) = − 5 z g ( z ) = z 5 + 1 g(z) = z^5 + 1 g ( z ) = z 5 + 1 z 5 − 5 z + 1 z^5 - 5z + 1 z 5 − 5 z + 1 ∣ z ∣ < 1 |z| \lt 1 ∣ z ∣ < 1 − 5 z -5z − 5 z z = 0 z = 0 z = 0
So exactly one root in ∣ z ∣ < 1 |z| \lt 1 ∣ z ∣ < 1
If you get this wrong, revise: Section 12.2 (Rouché’s Theorem).
Find the Möbius transformation that maps 1 ↦ 0 1 \mapsto 0 1 ↦ 0 i ↦ 1 i \mapsto 1 i ↦ 1 − 1 ↦ ∞ -1 \mapsto \infty − 1 ↦ ∞
Solution T ( z ) = ( z − 1 ) ( i − ( − 1 ) ) ( z − ( − 1 ) ) ( i − 1 ) = ( z − 1 ) ( i + 1 ) ( z + 1 ) ( i − 1 ) T(z) = \frac{(z - 1)(i - (-1))}{(z - (-1))(i - 1)} = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)} T ( z ) = ( z − ( − 1 )) ( i − 1 ) ( z − 1 ) ( i − ( − 1 )) = ( z + 1 ) ( i − 1 ) ( z − 1 ) ( i + 1 )
Simplify: i + 1 i − 1 = ( i + 1 ) ( − i − 1 ) ( i − 1 ) ( − i − 1 ) = − i 2 − 2 i − 1 − i 2 + 1 = − 2 i 2 = − i \frac{i + 1}{i - 1} = \frac{(i+1)(-i-1)}{(i-1)(-i-1)} = \frac{-i^2 - 2i - 1}{-i^2 + 1} = \frac{-2i}{2} = -i i − 1 i + 1 = ( i − 1 ) ( − i − 1 ) ( i + 1 ) ( − i − 1 ) = − i 2 + 1 − i 2 − 2 i − 1 = 2 − 2 i = − i
T ( z ) = − i ⋅ z − 1 z + 1 T(z) = -i \cdot \frac{z - 1}{z + 1} T ( z ) = − i ⋅ z + 1 z − 1
Verify: T ( 1 ) = 0 T(1) = 0 T ( 1 ) = 0 ✓ \checkmark ✓ T ( i ) = − i ⋅ i − 1 i + 1 = − i ⋅ ( − i ) = − 1 T(i) = -i \cdot \frac{i-1}{i+1} = -i \cdot (-i) = -1 T ( i ) = − i ⋅ i + 1 i − 1 = − i ⋅ ( − i ) = − 1
That gives − 1 -1 − 1 1 1 1
T ( z ) = ( z − z 1 ) ( z 2 − z 3 ) ( z − z 3 ) ( z 2 − z 1 ) T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} T ( z ) = ( z − z 3 ) ( z 2 − z 1 ) ( z − z 1 ) ( z 2 − z 3 ) z 1 = 1 z_1 = 1 z 1 = 1 z 2 = i z_2 = i z 2 = i z 3 = − 1 z_3 = -1 z 3 = − 1
T ( z ) = ( z − 1 ) ( i + 1 ) ( z + 1 ) ( i − 1 ) T(z) = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)} T ( z ) = ( z + 1 ) ( i − 1 ) ( z − 1 ) ( i + 1 )
T ( i ) = ( i − 1 ) ( i + 1 ) ( i + 1 ) ( i − 1 ) = 1 T(i) = \frac{(i - 1)(i + 1)}{(i + 1)(i - 1)} = 1 T ( i ) = ( i + 1 ) ( i − 1 ) ( i − 1 ) ( i + 1 ) = 1 ✓ \checkmark ✓
T ( 1 ) = 0 T(1) = 0 T ( 1 ) = 0 ✓ \checkmark ✓ T ( − 1 ) = ∞ T(-1) = \infty T ( − 1 ) = ∞ ✓ \checkmark ✓
So T ( z ) = ( z − 1 ) ( i + 1 ) ( z + 1 ) ( i − 1 ) T(z) = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)} T ( z ) = ( z + 1 ) ( i − 1 ) ( z − 1 ) ( i + 1 )
If you get this wrong, revise: Section 10.5 (Cross-Ratio).
Evaluate ∫ γ z 3 z 2 + 1 d z \int_\gamma \frac{z^3}{z^2 + 1}\, dz ∫ γ z 2 + 1 z 3 d z γ \gamma γ ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Solution z 3 z 2 + 1 \frac{z^3}{z^2 + 1} z 2 + 1 z 3 z = ± i z = \pm i z = ± i ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
At z = i z = i z = i R e s = i 3 2 i = − i 2 i = − 1 2 \mathrm{Res} = \frac{i^3}{2i} = \frac{-i}{2i} = -\frac{1}{2} Res = 2 i i 3 = 2 i − i = − 2 1 z = − i z = -i z = − i R e s = ( − i ) 3 − 2 i = i − 2 i = − 1 2 \mathrm{Res} = \frac{(-i)^3}{-2i} = \frac{i}{-2i} = -\frac{1}{2} Res = − 2 i ( − i ) 3 = − 2 i i = − 2 1
∫ γ z 3 z 2 + 1 d z = 2 π i ( − 1 2 − 1 2 ) = − 2 π i \int_\gamma \frac{z^3}{z^2 + 1}\, dz = 2\pi i\left(-\frac{1}{2} - \frac{1}{2}\right) = -2\pi i ∫ γ z 2 + 1 z 3 d z = 2 π i ( − 2 1 − 2 1 ) = − 2 π i
Alternatively: z 3 z 2 + 1 = z − z z 2 + 1 \frac{z^3}{z^2 + 1} = z - \frac{z}{z^2 + 1} z 2 + 1 z 3 = z − z 2 + 1 z ∫ γ z d z = 0 \int_\gamma z\, dz = 0 ∫ γ z d z = 0 ∫ γ z z 2 + 1 d z = 2 π i ( 1 / 2 + 1 / 2 ) = 2 π i \int_\gamma \frac{z}{z^2 + 1}\, dz = 2\pi i(1/2 + 1/2) = 2\pi i ∫ γ z 2 + 1 z d z = 2 π i ( 1/2 + 1/2 ) = 2 π i 0 − 2 π i = − 2 π i 0 - 2\pi i = -2\pi i 0 − 2 π i = − 2 π i ✓ \checkmark ✓
If you get this wrong, revise: Sections 8.4 and 8.5 (Residues).
Show that ∫ − ∞ ∞ cos 2 x x 2 + 1 d x = π e 2 \int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2} ∫ − ∞ ∞ x 2 + 1 c o s 2 x d x = e 2 π
Solution Consider ∫ − ∞ ∞ e 2 i x x 2 + 1 d x \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx ∫ − ∞ ∞ x 2 + 1 e 2 i x d x
f ( z ) = e 2 i z z 2 + 1 f(z) = \frac{e^{2iz}}{z^2 + 1} f ( z ) = z 2 + 1 e 2 i z z = i z = i z = i
R e s ( e 2 i z z 2 + 1 , i ) = e 2 i ⋅ i 2 i = e − 2 2 i \mathrm{Res}\!\left(\frac{e^{2iz}}{z^2 + 1}, i\right) = \frac{e^{2i \cdot i}}{2i} = \frac{e^{-2}}{2i} Res ( z 2 + 1 e 2 i z , i ) = 2 i e 2 i ⋅ i = 2 i e − 2
∫ − ∞ ∞ e 2 i x x 2 + 1 d x = 2 π i ⋅ e − 2 2 i = π e 2 \int_{-\infty}^{\infty} \frac{e^{2ix}}{x^2 + 1}\, dx = 2\pi i \cdot \frac{e^{-2}}{2i} = \frac{\pi}{e^2} ∫ − ∞ ∞ x 2 + 1 e 2 i x d x = 2 π i ⋅ 2 i e − 2 = e 2 π
Taking real parts: ∫ − ∞ ∞ cos 2 x x 2 + 1 d x = π e 2 \int_{-\infty}^{\infty} \frac{\cos 2x}{x^2 + 1}\, dx = \frac{\pi}{e^2} ∫ − ∞ ∞ x 2 + 1 c o s 2 x d x = e 2 π
If you get this wrong, revise: Section 9.7 (Fourier-Type Integrals).
Find the residue of f ( z ) = sin z z 4 f(z) = \frac{\sin z}{z^4} f ( z ) = z 4 s i n z z = 0 z = 0 z = 0
Solution sin z = z − z 3 / 6 + z 5 / 120 − ⋯ \sin z = z - z^3/6 + z^5/120 - \cdots sin z = z − z 3 /6 + z 5 /120 − ⋯
f ( z ) = z − z 3 / 6 + z 5 / 120 − ⋯ z 4 = 1 z 3 − 1 6 z + z 120 − ⋯ f(z) = \frac{z - z^3/6 + z^5/120 - \cdots}{z^4} = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots f ( z ) = z 4 z − z 3 /6 + z 5 /120 − ⋯ = z 3 1 − 6 z 1 + 120 z − ⋯
The coefficient of 1 / z 1/z 1/ z − 1 / 6 -1/6 − 1/6 R e s ( f , 0 ) = − 1 6 \mathrm{Res}(f, 0) = -\frac{1}{6} Res ( f , 0 ) = − 6 1
If you get this wrong, revise: Section 8.4 (Computing Residues).
Evaluate ∫ γ d z ( z − 1 ) 2 ( z − 2 ) \int_\gamma \frac{dz}{(z - 1)^2(z - 2)} ∫ γ ( z − 1 ) 2 ( z − 2 ) d z γ \gamma γ ∣ z − 1 ∣ = 1 / 2 |z - 1| = 1/2 ∣ z − 1∣ = 1/2
Solution Only z = 1 z = 1 z = 1 γ \gamma γ 2 2 2 z = 2 z = 2 z = 2
R e s ( f , 1 ) = d d z [ 1 z − 2 ] z = 1 = − 1 ( z − 2 ) 2 ∣ z = 1 = − 1 \mathrm{Res}(f, 1) = \frac{d}{dz}\left[\frac{1}{z - 2}\right]_{z=1} = -\frac{1}{(z-2)^2}\Big|_{z=1} = -1 Res ( f , 1 ) = d z d [ z − 2 1 ] z = 1 = − ( z − 2 ) 2 1 z = 1 = − 1
∫ γ f d z = 2 π i ⋅ ( − 1 ) = − 2 π i \int_\gamma f\, dz = 2\pi i \cdot (-1) = -2\pi i ∫ γ f d z = 2 π i ⋅ ( − 1 ) = − 2 π i
If you get this wrong, revise: Section 6.2 (CIF for Derivatives) and 8.4 (Residues).
Use the Cauchy-Riemann equations to show that f ( z ) = ∣ z ∣ 2 + 2 z ˉ f(z) = |z|^2 + 2\bar{z} f ( z ) = ∣ z ∣ 2 + 2 z ˉ f ′ ( z ) f'(z) f ′ ( z )
Solution f ( z ) = x 2 + y 2 + 2 x − 2 i y f(z) = x^2 + y^2 + 2x - 2iy f ( z ) = x 2 + y 2 + 2 x − 2 i y u = x 2 + y 2 + 2 x u = x^2 + y^2 + 2x u = x 2 + y 2 + 2 x v = − 2 y v = -2y v = − 2 y
u x = 2 x + 2 u_x = 2x + 2 u x = 2 x + 2 u y = 2 y u_y = 2y u y = 2 y v x = 0 v_x = 0 v x = 0 v y = − 2 v_y = -2 v y = − 2
CR: 2 x + 2 = − 2 ⇒ x = − 2 2x + 2 = -2 \Rightarrow x = -2 2 x + 2 = − 2 ⇒ x = − 2 2 y = 0 ⇒ y = 0 2y = 0 \Rightarrow y = 0 2 y = 0 ⇒ y = 0
f f f z = − 2 z = -2 z = − 2
f ′ ( − 2 ) = u x ( − 2 , 0 ) + i v x ( − 2 , 0 ) = ( 2 ( − 2 ) + 2 ) + 0 = − 2 f'(-2) = u_x(-2, 0) + iv_x(-2, 0) = (2(-2) + 2) + 0 = -2 f ′ ( − 2 ) = u x ( − 2 , 0 ) + i v x ( − 2 , 0 ) = ( 2 ( − 2 ) + 2 ) + 0 = − 2
If you get this wrong, revise: Section 3.1 (Cauchy-Riemann Equations).
Evaluate ∫ γ e z sin z ( z − π ) 3 d z \int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz ∫ γ ( z − π ) 3 e z s i n z d z γ \gamma γ ∣ z ∣ = 4 |z| = 4 ∣ z ∣ = 4
Solution Only z = π z = \pi z = π γ \gamma γ 3 3 3
By CIF for derivatives: ∫ γ f ( z ) ( z − π ) 3 d z = 2 π i 2 ! f ′ ′ ( π ) \int_\gamma \frac{f(z)}{(z - \pi)^3}\, dz = \frac{2\pi i}{2!}\,f''(\pi) ∫ γ ( z − π ) 3 f ( z ) d z = 2 ! 2 π i f ′′ ( π ) f ( z ) = e z sin z f(z) = e^z \sin z f ( z ) = e z sin z
f ′ ( z ) = e z sin z + e z cos z = e z ( sin z + cos z ) f'(z) = e^z \sin z + e^z \cos z = e^z(\sin z + \cos z) f ′ ( z ) = e z sin z + e z cos z = e z ( sin z + cos z ) f ′ ′ ( z ) = e z ( sin z + cos z ) + e z ( cos z − sin z ) = 2 e z cos z f''(z) = e^z(\sin z + \cos z) + e^z(\cos z - \sin z) = 2e^z \cos z f ′′ ( z ) = e z ( sin z + cos z ) + e z ( cos z − sin z ) = 2 e z cos z
f ′ ′ ( π ) = 2 e π cos π = − 2 e π f''(\pi) = 2e^\pi \cos\pi = -2e^\pi f ′′ ( π ) = 2 e π cos π = − 2 e π
∫ γ e z sin z ( z − π ) 3 d z = π i ⋅ ( − 2 e π ) = − 2 π i e π \int_\gamma \frac{e^z \sin z}{(z - \pi)^3}\, dz = \pi i \cdot (-2e^\pi) = -2\pi i\, e^\pi ∫ γ ( z − π ) 3 e z s i n z d z = π i ⋅ ( − 2 e π ) = − 2 π i e π
If you get this wrong, revise: Section 6.2 (Cauchy’s Integral Formula for Derivatives).
Problem. Evaluate ∮ ∣ z ∣ = 2 e z z − 1 d z \oint_{|z|=2} \frac{e^z}{z - 1} \, dz ∮ ∣ z ∣ = 2 z − 1 e z d z
Solution. The integrand has a simple pole at z = 1 z = 1 z = 1 e 1 = e e^1 = e e 1 = e ∮ ∣ z ∣ = 2 e z z − 1 d z = 2 π i ⋅ e = 2 π e i . \oint_{|z|=2} \frac{e^z}{z - 1} \, dz = 2\pi i \cdot e = 2\pi e i. ∮ ∣ z ∣ = 2 z − 1 e z d z = 2 π i ⋅ e = 2 π e i .
■ \blacksquare ■
Problem. Find the Taylor series of f ( z ) = 1 z f(z) = \frac{1}{z} f ( z ) = z 1 z 0 = 1 z_0 = 1 z 0 = 1
Solution. f ( z ) = 1 1 + ( z − 1 ) = ∑ n = 0 ∞ ( − 1 ) n ( z − 1 ) n f(z) = \frac{1}{1 + (z-1)} = \sum_{n=0}^{\infty} (-1)^n(z-1)^n f ( z ) = 1 + ( z − 1 ) 1 = ∑ n = 0 ∞ ( − 1 ) n ( z − 1 ) n ∣ z − 1 ∣ < 1 |z - 1| < 1 ∣ z − 1∣ < 1
■ \blacksquare ■
Complex analysis is the study of functions that respect the geometry of the complex plane. Unlike real functions, which can wiggle freely, holomorphic functions are astonishingly rigid: knowing their values on any small region determines them everywhere. The residue theorem encapsulates this rigidity by saying that contour integrals depend only on what singularities lie inside, not on the path taken. This is why complex analysis solves real integrals that resist elementary methods: the integral becomes a counting problem for poles, weighted by their residues.
Confusing complex conjugate and complex inverse. z ˉ = a − b i \bar{z} = a - bi z ˉ = a − bi z − 1 = z ˉ / ∣ z ∣ 2 = ( a − b i ) / ( a 2 + b 2 ) z^{-1} = \bar{z}/|z|^2 = (a - bi)/(a^2 + b^2) z − 1 = z ˉ /∣ z ∣ 2 = ( a − bi ) / ( a 2 + b 2 ) Fix: The conjugate is NOT the inverse; the inverse involves division by ∣ z ∣ 2 |z|^2 ∣ z ∣ 2 Wrong branch of the logarithm. log z = ln ∣ z ∣ + i ( arg z + 2 k π ) \log z = \ln|z| + i(\arg z + 2k\pi) log z = ln ∣ z ∣ + i ( arg z + 2 k π ) arg z ∈ ( − π , π ] \arg z \in (-\pi, \pi] arg z ∈ ( − π , π ] Fix: Always specify the branch when working with complex logarithms and powers.Cauchy’s theorem conditions. The function must be analytic on and inside the contour. Fix: If the function has singularities inside the contour, use the residue theorem instead. A[15_Problem Set] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
Cauchy-Riemann equations: u x = v y u_x = v_y u x = v y u y = − v x u_y = -v_x u y = − v x Cauchy’s integral theorem: ∮ γ f ( z ) d z = 0 \oint_\gamma f(z)\, dz = 0 ∮ γ f ( z ) d z = 0 f f f γ \gamma γ Residue theorem: ∮ γ f ( z ) d z = 2 π i ∑ Res ( f , z k ) \oint_\gamma f(z)\, dz = 2\pi i \sum \text{Res}(f, z_k) ∮ γ f ( z ) d z = 2 π i ∑ Res ( f , z k ) Taylor and Laurent series: power series representations; Laurent series include negative powers for singularities. Topic Site Link [Complex Numbers] A-Level View [Complex Numbers] IB View [Complex Numbers] University View
