Complex Numbers 12 Learning Outcomes In this chapter you have learned how to: Plot complex numbers on the Argand diagram Find the modulus of a complex number and calculate the complex conjugate Add, subtract, multiply and divide complex numbers Use complex numbers to perform transformations Solve equations with complex roots and use the Conjugate Root Theorem Write complex numbers in polar form Prove de Moivre’s Theorem Use de Moivre’s Theorem to evaluate (a + bi)n, a, b ∈ R, n ∈ Z Use de Moivre’s Theorem to find the roots of complex numbers

12 Complex Numbers Definition A Complex Number z is a number of the form 𝑧=𝑎+𝑏𝑖, 𝑎, 𝑏∈𝑅, 𝑖 2 =−1. a is called the real part of z, Re z . b is called the imaginary part of z, Im z . Examples: 2+3𝑖, 5−2𝑖, 1 2 + 3 4 𝑖, 2 −3𝑖 Note: The complex numbers 𝑧 1 =𝑎+𝑏𝑖 and 𝑧 2 =𝑐+𝑑𝑖 are equal if 𝑎=𝑐 and 𝑏=𝑑. Simplify 𝒊 𝟒 . 𝑖 4 =( 𝑖 2 ) 2 =(−1 ) 2 =1 𝑖=𝑖 𝑖 2 =−1 𝑖 3 =−𝑖 𝑖 4 =1 𝑖 5 =𝑖 𝑖 6 =−1 𝑖 7 =−𝑖 𝑖 8 =1 𝑖 9 =𝑖 Solve the equation 𝒛 𝟐 +𝟑𝟔=𝟎. 𝑧 2 +36=0 𝑧 2 =−36 𝑧=± −36 𝑧=± 36 −1 𝑧=±6𝑖 Note the repeating pattern.

12 Complex Numbers The Argand Diagram A complex number can be represented on a diagram called the Argand diagram. Here are the complex numbers 𝑧 1 =3+2𝑖, 𝑧 2 =4−3𝑖 and 𝑧 3 =−2−5𝑖 represented on an Argand diagram. -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im 𝑧 1 =3+2𝑖 𝑧 2 =4−3𝑖 𝑧 3 =−2−5𝑖 Modulus of a Complex Number Definition The modulus of a complex number, 𝑎+𝑏𝑖, is its distance from the origin. The modulus of 𝑎+𝑏𝑖 is written |𝑎+𝑏𝑖|. |𝒂+𝒃𝒊|= 𝒂 𝟐 + 𝒃 𝟐 If 𝒛 𝟏 =𝟕+𝟐𝟒𝒊 find | 𝒛 𝟏 |. 𝑧 1 = 7+24𝑖 = 7 2 + 24 2 = 49+576 = 625 =25

Addition, Subtraction, Multiplication and Division of Complex Numbers 12 Complex Numbers Addition, Subtraction, Multiplication and Division of Complex Numbers Addition 𝐈𝐟 𝒛 𝟏 =𝟐−𝟑𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟏𝟏+𝟓𝒊 evaluate 𝒛 𝟏 + 𝒛 𝟐 . 𝑧 1 + 𝑧 2 = 2−3𝑖 +(11+5𝑖) =2−3𝑖+11+5𝑖 Add the real parts to the real parts and the imaginary parts to the imaginary parts. 𝑧 1 + 𝑧 2 =13+2𝑖 Subtraction 𝐈𝐟 𝒛 𝟏 =𝟏𝟖+𝟏𝟔𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟏𝟒−𝟐𝒊 evaluate 𝒛 𝟏 − 𝒛 𝟐 . 𝑧 1 − 𝑧 2 = 18+16𝑖 −(14−2𝑖) =18+16𝑖−14+2𝑖 Add the real parts to the real parts and the imaginary parts to the imaginary parts. 𝑧 1 − 𝑧 2 =4+18𝑖

12 Complex Numbers Multiplication 𝐈𝐟 𝒛 𝟏 =𝟐−𝟕𝒊 𝐚𝐧𝐝 𝒛 𝟐 =𝟑+𝟕𝒊 evaluate 𝒛 𝟏 𝒛 𝟐 . 𝑧 1 𝑧 2 =(2−7𝑖)(3+7𝑖) =2(3+7𝑖)−7𝑖(3+7𝑖) =6+14𝑖−21𝑖−49 𝑖 2 =6+14𝑖−21𝑖−49(−1) =6+14𝑖−21𝑖+49 =55−7𝑖 Conjugate of a Complex Number Definition If 𝑧=𝑎+𝑏𝑖, then the 𝐜𝐨𝐧𝐣𝐮𝐠𝐚𝐭𝐞 of 𝑧 written as 𝑧 =𝑎−𝑏𝑖. Rule: Change the sign of the imaginary part. Example: If 𝑧=−2−12𝑖, then 𝑧 =−2+12𝑖.

12 Complex Numbers 𝟐+𝟏𝟏𝒊 𝟐+𝒊 . Division Calculate 𝟐+𝟏𝟏𝒊 𝟐+𝒊 . Note: When dividing by a complex number multiply the numerator and the denominator by the conjugate of the denominator. Step 1 Write down the conjugate of the denominator: 2+𝑖 =2−𝑖 Step 2 Multiply the denominator 2+𝑖 by 2−𝑖. 2+𝑖 2−𝑖 =2 2−𝑖 +𝑖(2−𝑖) =4−2𝑖+2𝑖− 𝑖 2 =5 Step 3 Multiply the numerator by 2−𝑖. 2+11𝑖 2−𝑖 =2 2−𝑖 +11𝑖(2−𝑖) =4−2𝑖+22𝑖−11 𝑖 2 =4+20𝑖+11 Step 4 2+11𝑖 2+𝑖 = 15+20𝑖 5 =15+20𝑖 =3+4𝑖

12 Complex Numbers Transformations The addition of complex numbers is the equivalent of a translation on the complex plane. If 𝒛 𝟏 =𝟐+𝟒𝒊 and 𝝎=𝟏+𝒊 plot 𝒛 𝟏 + 𝝎 on an Argand diagram. Describe the transformation that maps the addition of 𝒛 𝟏 to 𝒛 𝟏 +𝝎. The multiplication of a complex number by 𝒊 is the equivalent of rotating the complex number anti-clockwise through 𝟗𝟎 ° about the origin. 𝐈𝐟 𝒛 𝟏 =𝟐+𝒊 𝐚𝐧𝐝 𝝎=𝒊(𝟐+𝒊) describe the transformation that maps 𝒛 𝟏 to 𝝎. -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 Re Im 𝑧 1 +𝜔=3+5𝑖 𝑧 1 =2+4𝑖 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im 𝜔=1+𝑖 𝜔=−1+2𝑖 𝑧 1 =2+𝑖 𝑧 1 →𝑧 1 + 𝜔 is a translation. 𝑧 1 → 𝜔 is a+90° rotation.

Quadratic Equations with Complex Roots 12 Complex Numbers Quadratic Equations with Complex Roots 𝑎 𝑥 2 +𝑏𝑥+𝑐=0, 𝑎, 𝑏, 𝑐∈𝑅, 𝑎≠0 is solved by using the formula 𝑥= −𝑏± 𝑏 2 − 4𝑎𝑐 2𝑎 . In this formula, 𝑏 2 −4𝑎𝑐 is the discriminant. If 𝑏 2 −4𝑎𝑐<0, then the solutions (roots) will be complex. Solve the equation 𝒛 𝟐 −𝟔𝒛+𝟏𝟑=𝟎. -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Re Im 𝑧= 6± (−6 ) 2 −4(1)(13) 2(1) = 6± 36−52 2 𝑧=3+2𝑖 = 6± −16 2 = 6±4𝑖 2 𝑧=3−2𝑖 ∴𝑧=3±2𝑖 The Conjugate Root Theorem If the complex number 𝑧=𝑎+𝑏𝑖, 𝑎, 𝑏∈𝑅, is a root of a polymomial with real coefficients, then 𝑧 =𝑎−𝑏𝑖 (conjugate of z) is also a root.

Polynomials with Complex Roots 12 Complex Numbers Polynomials with Complex Roots If 𝟏+𝟐𝒊 is a root of 𝟑 𝒛 𝟑 −𝟓 𝒛 𝟐 +𝟏𝟑𝒛+𝟓, find the other roots. 1+2𝑖 is a root ∴1−2𝑖 is a root. (Conjugate Root Theorem) Step 1 Form a quadratic polynomial with these two roots. 𝑧 2 − sum of roots 𝑧+product of roots=0 Sum of roots= 1+2𝑖 + 1−2𝑖 =2 Product of roots= 1+2𝑖 1−2𝑖 =5 So the quadratic polynomial is 𝑧 2 −2𝑧+5=0 Step 2 Divide this function into the cubic function. Step 3 3𝑧+1 is a linear factor. Therefore, the solution to the equation 3𝑧+1 = 0 gives the real root. 3𝑧+1=0 ⟹𝑧= − 1 3 The three roots are 𝑧=1+2𝑖, 𝑧=1−2𝑖, 𝑧=− 1 3

√ 12 Complex Numbers Polar Form of a Complex Number (− 3 ) 2 +(1 ) 2 Re Im Definition 𝑟=|𝑥+𝑦𝑖| 𝑧=𝑥+𝑦𝑖 The polar form of a complex number z is 𝑟 𝑐𝑜𝑠𝜃+𝑖 𝑠𝑖𝑛𝜃 , where r is the modulus of the complex number and 𝜃 is its argument anticlockwise angle . tan 𝜃 = 𝑦 𝑥 ⟹𝜃= tan −1 𝑦 𝑥 𝑟 𝑦 𝜃 𝑥 Write the complex number − 𝟑 +𝒊 in polar form. Re Im 𝜃= tan −1 𝑦 𝑥 𝑟=|𝑥+𝑦𝑖| 𝑧=− 3 +𝑖 𝛼= tan −1 1 3 𝑟=|− 3 +𝑖| 1 𝜃 𝛼 𝑟= √ (− 3 ) 2 +(1 ) 2 𝛼= 𝜋 6 3 (0,0) 𝑟= 3+1 𝜃=𝜋− 𝜋 6 = 5𝜋 6 𝑟=2 Polar Form: 𝑟 cos𝜃+𝑖 sin𝜃 =2(cos 5𝜋 6 +𝑖 sin 5𝜋 6 )

12 Complex Numbers De Moivre’s Theorem Theorem If 𝑧=𝑟( cos 𝜃+𝑖 sin 𝜃) then 𝑧 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃), for 𝑛∈𝑍. i.e: [𝑟( cos 𝜃+𝑖 sin 𝜃) ] 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃), for 𝑛∈𝑍. The formal proof of De Moivre’s Theorem may be asked in the exam. Use de Moivre’s Theorem to write (𝟏+𝒊 ) 𝟏𝟎 in the form 𝒙+𝒚𝒊, 𝒙,𝒚∈𝑹. Step 1 Write 1+𝑖 in polar form, 𝑟 cos𝜃+𝑖 sin𝜃 . Re Im 𝜃= tan −1 𝑦 𝑥 𝜃= 𝜋 4 𝜃= tan −1 1 1+𝑖 Polar Form: 𝑟 cos𝜃+𝑖 sin𝜃 = 2 (cos 𝜋 4 +𝑖 sin 𝜋 4 ) 1 [𝑟( cos 𝜃+𝑖 sin 𝜃) ] 𝑛 = 𝑟 𝑛 ( cos 𝑛𝜃+𝑖 sin 𝑛𝜃) 𝜃 (0,0) 1 Step 2 Apply de Moivre’s Theorem. = ( 2 ) 10 [cos ( 𝜋 4 )+𝑖 sin ( 𝜋 4 ) ] 10 = ( 2 ) 10 [cos 10( 𝜋 4 )+𝑖 sin 10( 𝜋 4 )] = 2 5 cos ( 5𝜋 2 )+𝑖 sin ( 5𝜋 2 ) 𝑟=|𝑥+𝑦𝑖| 𝑟= (1 ) 2 +(1 ) 2 𝑟= 2 = 32[0+1𝑖] = 0+32𝑖

√ 12 Complex Numbers (2 ) 2 +(2 3 ) 2 If 𝝎=−𝟐−𝟐 𝟑 𝒊, solve the equation 𝒛 𝟒 −𝝎=𝟎. 𝑧=(−2−2 3 𝑖 ) 1 4 𝑧 4 =𝜔 𝑧 4 =−2−2 3 𝑖 Step 1 Express −2−2 3 𝑖 in general polar form: 𝑟[( cos 𝜃+2𝑛𝜋 + 𝑖 sin (𝜃+2𝑛𝜋)] 𝜔=−2−2 3 𝑖 𝜃 𝛼 𝛼= tan −1 2 3 2 ⟹𝛼= 𝜋 3 𝑟= √ (2 ) 2 +(2 3 ) 2 𝜃=−𝜋+ 𝜋 3 =− 2𝜋 3 𝑟= 16 =4 General polar form: 𝜔=4 cos − 2𝜋 3 +2𝑛𝜋 + 𝑖 sin − 2𝜋 3 +2𝑛𝜋 Step 2 Apply de Moivre’s Theorem: 𝑧=4 cos − 2𝜋 3 +2𝑛𝜋 + 𝑖 sin − 2𝜋 3 +2𝑛𝜋 1 4 =4 [ cos − 𝜋 6 + 𝑛𝜋 2 + 𝑖 sin (− 𝜋 6 + 𝑛𝜋 2 )] 1 4

12 Complex Numbers 𝑧=4 cos − 𝜋 6 + 𝑛𝜋 2 + 𝑖 sin − 𝜋 6 + 𝑛𝜋 2 . 1 4 Step 3 Find the four solutions of For 𝒏=𝟎 For 𝒏=𝟐 𝑧= 2 cos − 𝜋 6 + 𝑖 sin − 𝜋 6 𝑧= 2 cos − 𝜋 6 +𝜋 + 𝑖sin − 𝜋 6 +𝜋 = 2 3 2 − 1 2 𝑖 = 2 cos 5𝜋 6 + 𝑖sin 5𝜋 6 𝒛 𝟏 = 𝟔 𝟐 − 𝟐 𝟐 𝒊 = 2 − 3 2 + 1 2 𝑖 𝒛 𝟑 =− 𝟔 𝟐 + 𝟐 𝟐 𝒊 For 𝒏=𝟏 For 𝒏=𝟑 𝑧= 2 cos − 𝜋 6 + 3𝜋 2 + 𝑖sin − 𝜋 6 + 3𝜋 2 𝑧= 2 cos − 𝜋 6 + 𝜋 2 + 𝑖sin − 𝜋 6 + 𝜋 2 = 2 cos 𝜋 3 + 𝑖sin 𝜋 3 = 2 cos 4𝜋 3 + 𝑖sin 4𝜋 3 = 2 1 2 + 3 2 𝑖 𝒛 𝟐 = 𝟐 𝟐 − 𝟔 𝟐 𝒊 = 2 − 1 2 − 3 2 𝑖 𝒛 𝟒 =− 𝟐 𝟐 − 𝟔 𝟐 𝒊