Ch. 2- Complex Numbers>Background Chapter 2: Complex Numbers I. Background • Complex number (z): z = A + iB where i 1 (note: engineers use “real part”: Re{z} = A “imaginary part”: Im{z} = B (both A and B are real numbers) • Graphical Representation: ex: 2+3i 3-4i j 1 ) Ch. 2- Complex Numbers>Background • Polar vs. rectangular coordinates: Rectangular coordinates: z = A + iB Polar coordinates: z = reiθ (note: |z| ≡ r , z = θ) To move between the two: A + iB = (r cosθ) + i (r sinθ) = reiθ because eiθ = cosθ + sinθ (see section 7-9 for proof) so: 2 2 A r cos r A B or 1 B B r sin tan A θ is in radians, r ≥ 0 ex: z=3-i4 (write in polar coordinates) Ch. 2- Complex Numbers>Background • Beware when using tan 1 ( BA ) tan 1 ( BA ) tan 1 ( BA ) Your calculator can’t tell the difference between θ and θ+π. It should always find 2 0 2 . You may need to add π to get to the correct quadrant! ex: z = -3 + i4 In practice, polar form reiθ is often easier to use because it’s easier to differentiate and integrate reiθ. Ch. 2- Complex Numbers>Algebra>Complex Conjugate II. Algebra with complex numbers 1) Complex conjugate: Let z = A + iB Then z (or z*) A iB is the complex conjugate. In polar coordinates: z = reiθ z* = re-iθ Why? Recall: z = reiθ = r (cosθ + isinθ) z* = r (cosθ - isinθ) = r (cos(-θ) + isin(-θ)) = re-iθ ex: 1) z = 2 + i3 2) z = 3 – i4 3) z = 5ei(/3) Note: z z* = |z|2 (Proof in a minute…) Ch. 2- Complex Numbers>Algebra>Addition For the remaining examples: Let z1 = 2 + i3 r = 3.6, (θ) = 0.98 rad z1 = 3.6ei(0.98) z2 = 4 – i2 r = 4.5, (θ) = -0.46 rad z1 = 4.5ei(-0.46) 2) Addition: (A + iB) + (C + iD) = (A+C) + i(B+D) i i there is no easy way to do this in polar coordinates: r1 e r2 e ? 1 ex: z1 + z2 (in rectangular coordinates) ex: z1 + z2 (in polar coordinates) 2 Ch. 2- Complex Numbers>Algebra>Multiplication 3) Multiplication: (A + iB) (C + iD) = AC + i(AB) + i(BC) – BD (term by term) ( r1e ex: z1•z2 (rectangular) ex: z1•z2 (polar) i 1 )( r2 e i 2 ) ( r1 r2 ) e i ( 1 2 ) Ch. 2- Complex Numbers>Algebra>Multiplying by Complex Conjugate 4) Multiplying by complex conjugate: ex: z1z1* |z1|2 This is true for any complex number: |z1|2 = z1z1* Proof: Let z = A + iB z z*= (A + iB) (A – iB) = A2 – iAB + iBA + B2 = A2 + B2 = r2 = |z|2 Ch. 2- Complex Numbers>Algebra>Division 5) Division: r1 e r2 e i 1 i 2 r1 r2 e i ( 1 2 ) (more difficult in rectangular form) ex: ex: z1 z2 z1 z2 (polar) (rectangular) Ch. 2- Complex Numbers>Algebra>More Complex Conjugates & Complex Equations 6) More complex conjugates: Say I want the complex conjugate of a messy equation: 2 i 3 x ix 2 z3 9 x ix 2 3 4 Change all i -i 2 i 3 x ix 2 z3* 9 x ix 2 3 4 7) Complex equations: (A + iB) = (C + iD) ex: z1 + z2 = x + i(3x + y) iff A = C and B = D Ch. 2- Complex Numbers>Algebra>Powers 8) Powers: do these in polar form: ( r1 e i 1 n ) r1 e n rectangular form switch to polar first ex: z12 (polar) ex: z12 (rectangular) in 1 9) Roots: Polar coordinates: ex: ( re i 1 n 1 n ) r e i ( n ) z1 Check by changing back to rectangular coordinates. Find another root (add 2 to ). Convert back to rectangular coordinates: Ch. 2- Complex Numbers>Algebra>Roots In general: z 1 n ex: z 4 ex: z 27 Ch. 2- Complex Numbers>Algebra>Roots has n possible roots! z 2 find 3 z Ch. 2- Complex Numbers>Algebra>Complex Exponentials 10) Complex Exponentials: let z = x + iy then ez = ex+iy = exeiy = ex(cosy + isiny) ex: e e e e 3 i 3 i 2 i i 2 Ch. 2- Complex Numbers>Algebra>Trig Functions 11) Trig Functions: e e e e (e e i cos i sin i i cos i sin e i 2 cos cos i e i 2 cos i sin i i i e cos i sin ) e i 2 i sin sin e i e 2i i Ch. 2- Complex Numbers>Algebra>Trig Functions This is very useful for derivatives and integrals: ex: d dz (cos z ) Good Trick: 1 i i i 1 i ex: cos( 2 x ) cos( 3 x ) dx Ch. 2- Complex Numbers>Algebra>Hyperbolic Functions 12) Hyperbolic Functions: iz iz sin z e e 2i cos z e e 2 iz iz Usual sin/cos functions Likewise: tanh z sinh z cosh z sech z 1 cosh z etc… ↔ sinh z cosh z z e e 2 z z e e 2 z Hyperbolic functions (entirely real) Ch. 2- Complex Numbers>Algebra>Natural Logarithm 13) Natural logarithm: i i ln z ln( re ) ln( r ) ln( e ) ln( r ) i Note: i = i(+2) = i(+4) = … So ln(z) has infinitely many solutions: ln(z) = ln(r) + i = ln(r) + i(+2) = … ‘Principal solution’ has 0 , 2 . Ch. 2- Complex Numbers>Example: RLC Circuit Physics Example: RLC Circuit applied emf V V sin ωt Then I I sin (ωω ) Find I0 & Φ and Z (the complex impedance). From Physics 216: V R IR RI 0 sin( t ) VL L dI dt VC 1 C LI 0 cos( t ) Idt 1 C I 0 cos( t ) What is the impedance? Z V I V R VC V L I (yuck!) Ch. 2- Complex Numbers>Example: RLC Circuit It’s easier to do this: V V0e i t Ch. 2- Complex Numbers>Example: RLC Circuit What’s the physically real I? i ( t ) I I 0e where V0 I0 R ( L 2 tan 1 1 C 1 C ) 2 L R Recall our actual driving voltage: We wrote this as V 0 e Only Im V 0 e i t i t . has any physical reality. So, same goes for current: The physically real current is Im I 0 e i ( t ) I 0 sin( t ) with same I o & Φ as above. Ch. 2- Complex Numbers>Example: RLC Circuit Resonance Defn: The frequency at which Z is entirely real: Z R i L 1 C So, at resonance ωR: RL 1 RC 0 R 1 LC (resonance frequency) Note that for this circuit, I0 is max when ω=ωR: I0 V0 R ( L 2 See plot of I0 vs. ω. 1 C ) 2 V0 R (at resonance) Ch. 2- Complex Numbers>Example: RLC Circuit Complex Impedances We found: V R RI XR R V L i LI V L X L I X L i L VC 1 i C I VC X C I where XC 1 i C “reactance” or “complex impedance” Complex impedances behave just like resistors in series and parallel. Ch. 2- Complex Numbers>Example: RLC Circuit From Physics 216: Wave : Asin( ω - kx φ) Wavelength : k Period : ω 2π T 2π λ λ T 2π k 2π ω Phase : φ Velocity : v Amplitude ω k (to the right) : A Instead, it’s easier to use complex notation: Ae i( ω( kx φ)