Questions and Answers #15 Complex Numbers

Questions and Answers Sheet 15 Complex Numbers Question #1 How do you simplify −3𝑖(2 − 5𝑖) − (7 + 𝑖)? Answer: Step 1 −3𝑖(2 − 5𝑖) − (7 + 𝑖) = −6𝑖 + 15𝑖 2 − 7 − 𝑖 = −7𝑖 + 15(−1) − 7 = −7𝑖 − 15 − 7 = −7𝑖 − 22 or, −(22 + 7𝑖) Question #2 How do you simplify 𝑖 6 (2𝑖 − 𝑖 2 − 3𝑖 3 )? Answer: Step 1 Expand expresion. 𝑖 6 (2𝑖 − 𝑖 2 − 3𝑖 3 ) = 2𝑖 7 − 𝑖 8 − 3𝑖 9 But we know that: 𝑖1 = 𝑖 𝑖 2 = −1 (by definition) 𝑖 3 = 𝑖 2 ⋅ 𝑖 = −𝑖 𝑖 4 = 𝑖 2 ⋅ 𝑖 2 = (−1) ⋅ (−1) = 1 𝑖 5 = 𝑖 4 ⋅ 𝑖 = 1 ⋅ 𝑖 = 𝑖 and starts again 𝑖 6 = 𝑖 5 ⋅ 𝑖 = 𝑖 ⋅ 𝑖 = 𝑖 2 = −1 𝑖 7 = 𝑖 6 ⋅ 𝑖 = −𝑖 for this cliclic result, we can write: 𝑖 6 (2𝑖 − 𝑖 2 − 3𝑖 3 ) = 2𝑖 7 − 𝑖 8 − 3𝑖 9 = −2𝑖 − 1 − 3𝑖 = −1 − 5𝑖 Question #3 How do you change 10𝑧12 into polar coordinates where 𝑧 = 2 + 4𝑖 ? Answer: 2 + 4𝑖 𝑏 𝜃 = arctan ( ) 𝑎 𝜃 = arctan(2) 𝑟 = √22 + 42 = √20 = 2√5 polar form of 2 + 4𝑖 ⇒ 2√5(cos(arctan(2)) + 𝑖 sin(arctan(2))) 12 (2√5(cos(arctan(2)) + 𝑖 sin(arctan(2)))) using the moivre formula 12 (2√5) (cos(arctan(2)) + 𝑖 sin(arctan(2)))12 6,4 ⋅ 107 (cos(12 arctan(2)) + 𝑖 sin(12 arctan(2))) = 𝑧12 6,4 ⋅ 108 (cos(12 arctan(2)) + 𝑖 sin(12 arctan(2))) = 10𝑧12 Question #4 How do you write the complex number 𝑧 = 8 − 8𝑖 in trigonometric form? Answer: 8 − 8𝑖 = 8(1 − 𝑖) = 8√2 ( 1 √2 − 𝑖 √2 ) 𝜋 = 8√2𝑒 −𝑖 4 Question #5 How do you write the complex number 𝑧 = −8 in trigonometric form? Answer: 𝜋 𝜋 8 (cos ( ) + 𝑖 sin ( )) 2 2 Explanation: 𝑧 = −8 + 0𝑖 In general 𝑎 + 𝑏𝑖 ⇒ 𝑟(cos(𝜃) + 𝑖 sin(𝜃)) with 𝑟 = √𝑎2 + 𝑏 2 and 𝜃 = b  arctan( a )if a  0    if a = 0and b  0 2   + arctan( b )if a  0  a  −  if a = 0 and b  0  2 undefined if a=0 and b = 0  Question #6 How do you write the complex number 𝑧 = 6(cos 4 5∘ + 𝑖 sin 4 5∘ ) in standard form? Answer: I found: 3√2 + 3√2𝑖 Explanation: We need to evaluate the values of sin and cos and multiply by 6: 𝑧 = 6(cos(45∘ ) + 𝑖 sin(45∘ )) = 6 ( 3 =6 ( √2 √2 +𝑖 )= 2 2 2 2 +i = 3 2 + 3 2i 2 2 Question #7 How do you find the quotient of the complex numbers and express your answer in trigonometric form 5(cos(15∘ )+𝑖 sin(15∘ )) 3(cos(70∘ )+𝑖 sin(70∘ )) ? Answer: 5 ( ) (cos 5 5∘ − 𝑖 sin 5 5∘ ) . 3 Explanation: Use 𝑒 ±𝑖𝜃 = cos 𝜃 ± 𝑖 sin 𝜃 . 5 𝑒 𝑖15 The ratio = ( ) 3 5 ∘ ∘ 𝑒 𝑖70 ∘ = ( ) 𝑒 𝑖(15−70) 3 5 = ( ) (cos 5 5∘ − 𝑖 sin 5 5∘ ) 3 Question #8 How do you express the complex number in standard form: √3 ( cos 𝜋 +𝑖 6 sin 𝜋 6 )? Answer: Step 1 𝜋 √3 6 2 cos ( ) = 𝜋 1 6 2 sin ( ) = The complex number is 𝜋 𝜋 6 6 𝑧 = √3 (cos ( ) + 𝑖 sin ( )) Step 2 𝑧 = √3 ( 3 √3 2 2 = + √3 2 1 + 𝑖) 2 𝑖 Question #9 How do you convert the complex the number into polar representation: 2√3 + 2𝑖 ? Answer: 2√3 + 2𝑖 = 4( √3 2 𝑖 + ) 2 𝜋 𝜋 6 6 = 4 (cos ( ) + 𝑖 sin ( )) 𝜋 = 4𝑒 𝑖 6 Question #10 How do you express the complex number in standard form: 8(cos 3 0 + 𝑖 sin 3 0) ? Answer: 8(cos 3 0∘ 𝑖 sin 3 0∘ ) = 8( √3 2 1 +𝑖× ) 2 = 4√3 + 4𝑖 Question #11 How do you express the complex number in rectangular form: 𝑧 = 5(cos 1 20∘ + 𝑖 sin 1 20∘ ) ? Answer: 5 5 Answer: (− , √3) 2 2 Using the formulae that links Polar to Rectangular coordinates. −𝑥 = 𝑟 cos 𝜃 −𝑦 = 𝑟 sin 𝜃 here 𝑟 = 5 and 𝜃 = 120∘ 1 5 2 2 Hence: 𝑥 = 5 cos 1 20∘ = 5 × − = − and 𝑦 = 5 sin 1 20∘ = 5 × √3 2 5 = √3 2 Question #12 How do you express the complex number in trigonometric form: 1 − (√3)𝑖 ? Answer: to convert to trigonometric form that is 𝑟(cos 𝜃 + 𝑖 sin 𝜃) where −𝑟 = √𝑥 2 + 𝑦 2 𝑦 −𝜃 = tan−1 ( ) − 𝜋 < 𝜃 ≤ 𝜋 𝑥 Here 𝑥 = 1 and 𝑦 = −√3 2 ⇒ 𝑟 = √12 + (−√3) = 2 𝜃 = tan−1 (−√3) = − 𝜋 3 𝜋 𝜋 3 3 ⇒ 2 (cos (− ) + 𝑖 sin (− )) 𝜋 𝜋 3 3 = 2 (cos ( ) − 𝑖 sin ( )) 𝜋 𝜋 3 3 ⇒ 1 − √3𝑖 = 2 (cos ( ) − 𝑖 sin ( )) Question #13 How do you express the complex number in rectangular form: 12(cos 6 0∘ + 𝑖 sin 6 0∘ ) ? Answer: Answer: 6 + 𝑖6√3; [𝑎 + 𝑖𝑏] Explanation: Complex number; 12(cos 6 0 + 𝑖 sin 6 0) 1 √3 2 2 Rectangular form: 12 ⋅ + 12 ⋅ 𝑖 or Therefore, 6 + 6√3𝑖 or 6 + 𝑖6√3; [𝑎 + 𝑖𝑏] Question #14 How do you write the trigonometric form of √3 + 𝑖? Answer: 𝜋 𝜋 Answer: √3 + 𝑖 = 2 cos ( ) + 2𝑖 sin ( ) 6 6 Explanation: A number 𝑎 + 𝑖𝑏 can be written in trigonometric form as 𝑟 cos 𝜃 + 𝑖𝑟 sin 𝜃 As 𝑟 cos 𝜃 = 𝑎 and 𝑟 sin 𝜃 = 𝑏 , squaring and adding them we get 𝑟 2 = 𝑎2 + 𝑏 2. 2 As such for √3 + 𝑖, 𝑟 = √(√3) + 12 = √3 + 1 = √4 = 2 and cos 𝜃 = 𝑖. 𝑒. 𝜃 = √3 2 and sin 𝜃 = 1 2 𝜋 6 𝜋 𝜋 Hence in trigonometric form √3 + 𝑖 = 2 cos ( ) + 2𝑖 sin ( ) 6 6 Question #15 How do you write the trigonometric form of −1 + √3𝑖 ? Answer: To convert from cartesian to trig. form that is (𝑥, 𝑦) → 𝑟(cos 𝜃 + 𝑖 sin 𝜃) 𝑟 = √𝑥 2 + 𝑦 2 𝑦 𝜃 = tan−1 ( ) 𝑥 here 𝑥 = −1and � = √3 2 ⇒ 𝑟 = √(−1)2 + (√3) = 2 −1 + 𝑖√3 is in the second quadrant so we must ensure that 𝜃 is in the second quadrant 𝜃 = tan−1 (√3) = 𝜋 3 ← related acute angle 𝜋 2𝜋 3 3 ⇒ 𝜃 = (𝜋 − ) = ← in second quadrant 2𝜋 2𝜋 3 3 ⇒ −1 + 𝑖√3 = 2 (cos ( ) + 𝑖 sin ( )) Question #16 5 How do you write the trigonometric form of 2 (√3 − 𝑖) ? Answer: 𝜋 𝜋 6 6 Answer: 5 (cos (− ) + 𝑖 sin (− )) Explanation: The trigonometric form of a complex number 𝑧 = 𝑎 + 𝑖𝑏 is 𝑧 = |𝑧|(cos 𝜃 + 𝑖 sin 𝜃) 𝑎 where cos 𝜃 = |𝑧| 𝑏 and sin 𝜃 = |𝑧| 5 Here 𝑧 = (√3 − 𝑖) 2 5 5 |𝑧| = √3 + 1 = ⋅ 2 = 5 2 2 𝑧 = 5( √3 2 1 − 𝑖) 2 √3 cos 𝜃 = 2 sin 𝜃 = − 1 2 𝜋 𝜃 = − , [𝑚𝑜𝑑2𝜋] 6 The trigonometric form is 𝜋 𝜋 6 6 𝜋 𝑧 = 5 (cos (− ) + 𝑖 sin (− )) = 5𝑒 −𝑖 6 Question #17 7𝜋 7𝜋 √2 4 4 2 How do you simplify [√2 (cos ( ) + 𝑖 sin ( ))] ÷ [ and express the result in rectangular form? Answer: Step 1 7𝜋 7𝜋 √2(cos( 4 )+𝑖 sin( 4 )) 3 3𝜋 √2 (cos( )+𝑖 sin( )) 2 𝜋 4 7𝜋 = 7𝜋 2(cos( 4 )+𝑖 sin( 4 )) 3𝜋 3𝜋 (cos( )+𝑖 sin( )) 4 4 Step 2 3𝜋 3𝜋 4 4 (cos ( ) + 𝑖 sin ( ))] 7𝜋 = 𝑖 2𝑒 4 3𝜋 𝑖 𝑒 4 7𝜋 3𝜋 = 2𝑒 𝑖( 4 − 4 ) = 2𝑒 𝑖𝜋 = 2(cos 𝜋 + 𝑖 sin 𝜋) = 2(−1 + 𝑖 ⋅ 0) = −2 + 𝑖 ⋅ 0 Question #18 7𝜋 7𝜋 11𝜋 6 6 3 How do you simplify 20 (cos ( ) + 𝑖 sin ( )) ÷ 15 (cos ( ) + 𝑖 sin ( 11𝜋 3 )) and express the result in rectangular form? Answer: In polar complex, that's: 7𝜋 20𝑒 𝑖 6 15𝑒 𝑖 11𝜋 3 4 5𝜋 = 𝑒− 2 3 𝜋 4 = 𝑒 −2 3 =− 4𝑖 3 Question #19 3𝜋 3𝜋 𝜋 𝜋 4 4 2 2 How do you simplify 2 (cos ( ) + 𝑖𝑠𝑖𝑛 ( )) ⋅ √2 (cos ( ) + 𝑖𝑠𝑖𝑛 ( )) and express the result in rectangular form? Answer: 3𝜋 3𝜋 𝜋 𝜋 2 (cos ( ) + 𝑖 sin ( )) ⋅ √2 (cos ( ) + 𝑖 sin ( )) 4 4 2 2 3𝜋 𝜋 3𝜋 𝜋 𝜋 3𝜋 3𝜋 3𝜋 𝜋 4 2 4 2 2 4 4 4 2 3𝜋 𝜋 3𝜋 𝜋 3𝜋 𝜋 𝜋 3𝜋 4 2 4 2 4 2 2 4 = 2√2{cos ( ) cos ( ) + 𝑖 cos ( ) sin ( ) + 𝑖 cos ( ) sin ( ) + 𝑖 2 sin ( ) sin ( ) sin ( )} = 2√2{cos ( ) cos ( ) − sin ( ) sin ( ) + 𝑖 (cos ( ) sin ( ) + 𝑖 cos ( ) sin ( ))} = 2√2{cos ( 3𝜋 4 𝜋 3𝜋 2 4 + ) + 𝑖 sin ( 5𝜋 5𝜋 4 4 = 2√2 (cos ( ) + 𝑖 sin ( )) 𝜋 𝜋 4 4 2√2 (− cos ( ) − 𝑖 sin ( )) = 2√2 (− 1 √2 = −2 − 2𝑖 −𝑖 1 √2 ) 𝜋 + )} 2 Question #20 How do I find the trigonometric form of the complex number 3 − 3√3𝑖. Answer: In trigonometric form: 6(cos 5. 236 + 𝑖 sin 5. 236) Explanation: Let 𝑍 = 𝑎 + 𝑖𝑏; 𝑍 = 3 − 3√3𝑖; 𝑎 = 3, 𝑏 = −3√3; Z is in 4-th quadrant. Modulus |𝑍| = √𝑎2 + 𝑏 2 2 = (√32 + (−3√3) ) = √36 = 6 tan  =| b 3 3 |= or tan  = 3 a 3 𝛼 = tan−1 (√3) ≈ 1.0472 𝜃 is on 4-th quadrant 𝜃 = 2𝜋 − 1.0472 ≈ 5.236 Argument, 𝜃 = 5.236 . In trigonometric form expressed as |𝑍|(cos 𝜃 + 𝑖 sin 𝜃) = 6(cos 5. 236)