Questions and Answers #10 Complex Numbers

Questions and Answers Sheet 10 Complex Numbers Question #1 𝜋 𝜋 12 12 How do you find the cube root of 81 (cos ( ) + 𝑖 sin ( )) ? Answer: Step 1 4 4 4 33 (cos 5∘ + 𝑖 sin 5∘ ), 33 (cos 1 25∘ + 𝑖 sin 1 25∘ )and 33 (cos 2 45∘ + 𝑖 sin 2 45∘ ) . Explanation:   1 (81(cos( ) + i sin( )) 3 12 12 1 3 = 81 1 𝜋 3 (𝑐𝑖𝑠 ) 12 1  3 (cis (( (2k + )), k = 0,1, 2 3 12 4 3 4 = 33 (𝑐𝑖𝑠5∘ , 𝑐𝑖𝑠125∘ , 𝑐𝑖𝑠245∘ ), using 𝜋 = 180∘ 4 = 33 (cos 5∘ + 𝑖 sin 5∘ ) Step 2 4 33 (cos 1 25∘ + 𝑖 sin 1 25∘ ) and 4 33 (cos 2 45∘ + 𝑖𝑠𝑖𝑛245∘ ) . Question #2 𝜋 𝜋 4 4 How do you find the 4th root of 2 (cos ( ) + 𝑖 sin ( )) ? Answer: The principal 4th root is: 4 𝜋 𝜋 𝜋 4 𝜋 √2 (cos ( ) + 𝑖 sin ( )) = √2 (cos ( ) + 𝑖 sin ( )) 4 4 16 16 There are 3 other 4-th roots. Explanation: In general, if n is a positive integer and 𝜃 ∈ (−𝜋, 𝜋] then the principal n-th root is given by: 𝑛 𝑛 𝜃 𝜃 √𝑟(cos 𝜃 + 𝑖 sin 𝜃) = √𝑟 (cos ( ) + 𝑖 sin ( )) 𝑛 𝑛 So in our example: 4 𝜋 𝜋 𝜋 4 𝜋 √2 (cos ( ) + 𝑖 sin ( )) = √2 (cos ( ) + 𝑖 sin ( )) 4 4 16 16 Note that the primitive Complex 4-th root of 1 is i, so the other 4-th roots are: 𝜋 𝜋 4 𝜋 𝜋 4 7𝜋 7𝜋 4 𝜋 𝜋 4 3𝜋 3𝜋 4 4 5𝜋 5𝜋 𝑖 √2 (cos ( ) + 𝑖 sin ( )) = √2 (cos ( ) + 𝑖 sin ( )) 16 16 16 16 𝑖 2 √2 (cos ( ) + 𝑖 sin ( )) = √2 (cos (− ) + 𝑖 sin (− )) 16 16 16 16 𝑖 3 √2 (cos ( ) + 𝑖 sin ( )) = √2 (cos (− ) + 𝑖 sin (− )) 16 16 16 16 Question #3 How do you simplify 𝑧 6 = 𝑎 + 𝑏𝑖 given 𝑧 = √3 + 𝑖 ? Answer: Step 1 If you are familiar with Euler's rule 𝑧 = 𝑒 𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥 we can express z as 𝑧 = 2( √3 2 𝑖 cos 𝜋 2 6 + ) = 2( +𝑖 sin 𝜋 6 𝜋 ) = 2𝑒 𝑖 6 𝜋 6 thus 𝑧 6 = (2𝑒 𝑖 6 ) 𝜋 6 = 26 × (𝑒 𝑖 0 ) = 64 × 𝑒 𝑖𝜋 and then reversing back into original form = 64(cos 𝜋 + 𝑖 sin 𝜋) = 64(−1 + 𝑖(0)) = −64 Question #4 Find the fourth roots of −17𝑖? Answer: Step 1 Let us wrote −17𝑖 in polar form and as its modulus is 17 and number is 117 × (−𝑖),we can write it as 3𝜋 −17𝑖 = 17𝑒 2 𝑖 4 Let 𝑥 = √−17𝑖 , then 3𝜋 𝑥 4 = −17𝑖 = 17𝑒 2 𝑖 3𝜋 Hence 𝑥 = √17𝑒 8 𝑖 4 Other roots can be obtained by adding 7𝜋 i.e. √17𝑒 8 𝑖 , √17𝑒 4 4 11𝜋 𝑖 8 4 and √17𝑒 𝜋 2 to 3𝜋 8 in succession (as they are one-fourth of 2𝜋 ) 15𝜋 𝑖 8 To find them in the form of 𝑎 + 𝑏𝑖 , you can use the following 𝑟𝑒 𝑖𝜃 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 Question #5 z is a complex number, 𝑧 = 𝑎 + 𝑖𝑏 and (2 + 𝑖)𝑧 = 1 − 𝑧 , what is z in terms of i? Answer: Step 1 (2 + 𝑖)𝑧 = 1 − 𝑧 ⇒ (2 + 𝑖)𝑧 + 𝑧 = 1 ⇒ (2 + 𝑖 + 1)𝑧 = 1 ⇒ (3 + 𝑖)𝑧 = 1 ⇒𝑧= 1 3+𝑖 3−𝑖 = (3+𝑖)(3−𝑖) = = 3−𝑖 10 3 10 − ∴𝑧= 1 10 3 10 𝑖 − 1 10 𝑖 Question #6 How do you solve 6𝑥 2 − 5𝑥 + 3 = 0? Answer: Step 1 𝑦 = 6𝑥 2 − 5𝑥 + 3 = 0 𝐷 = 𝑑 2 = 𝑏 2 − 4𝑎𝑐 = 25 − 72 = −47 < 0 There are no real roots, meaning the parabola graph doesn't intersect the x-axis. There are 2 complex roots: 𝑑 = ±√47𝑖 5 𝑥= 12 ± √47𝑖12 Question #7 What is the square root of 2i ? Answer: Step 1 √2𝑖 = {1 + 𝑖, − 1 − 𝑖} Let us look at some details. Let 𝑧 = √2𝑖 (Note that z are complex numbers.) by squaring. ⇒ 𝑧 2 = 2𝑖 by using the exponential form z=re^{i\theta} 𝜋 ⇒ 𝑟 2 𝑒 𝑖(2𝜃) = 2𝑖 = 2𝑒 𝑖( 2 +2𝑛𝜋) r 2 = 2  r = 2     2 = + 2n   = + n  2 4 𝜋 So, 𝑧 = √2𝑒 𝑖( 4 +𝑛𝜋) by Eular's Formula: 𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 𝜋 𝜋 4 4 ⇒ 𝑧 = √2 [cos ( + 𝑛𝜋) + 𝑖 sin ( + 𝑛𝜋)] = √2 (± 1 √2 ± 1 √2 𝑖) = ±1 ± 𝑖 I kept the following original post just in case someone needs it. 1 1 1 (2𝑖)2 = (2)2 (𝑖)2 1 (𝑖)2 = −1 1 1 (2𝑖)2 = (2)2 × −1 1 (2)2 = 1.41 1 (2𝑖)2 = 1.41 × −1 = −1.41 Question #8 How do you find the square root of 5(cos 1 20 + 𝑖 sin 1 20)? Answer: Step 1 Use cos(180∘ + 𝑎) = −𝑐𝑜𝑠 and sin(180∘ + 𝑎) = − sin 𝑎 1 (5(cos 1 20∘ + 𝑖 sin 1 20∘ ))2 1 √5(cos(2𝑘(180)∘ + 120∘ ) + 𝑖 sin(2𝑘(180)∘ + 120∘ ))2 = √5 (cos ( 2𝑘(180)+120∘ 2 ) + 𝑖 sin ( = √5(cos 6 0∘ + 𝑖 sin 6 0∘ ) = √5 2 (2𝑘(180)∘ +120∘ ) 2 )) × 𝑘 = 0, 1 (1 + 𝑖√3), for 𝑘 = 0 and = √5(cos 2 40∘ + 𝑖 sin 2 40∘ ) for 𝑘 = 1 = 5(cos(180 + 60 ) + i sin(180 + 60 ) = −√5(cos 6 0∘ + 𝑖 ∈ 60∘ ) =− √5 2 (1 + 𝑖√3) Question #9 How do you find the square root of 16(cos 6 0∘ + 𝑖 sin 6 0∘ )? Answer: Step 1 𝑧 = 16(cos 6 0∘ + 𝑖 sin 6 0∘ ) 1 √𝑧 = 𝑧 2 = √16(cos 6 0∘ + 𝑖 sin 6 0∘ ) 1 1 𝑧 2 = 4(cos 6 0 + 𝑖 sin 6 0)2 now using DeMoivre's theorem (cos 𝜃 + 𝑖 sin 𝜃)𝑛 = cos 𝑛 𝜃 + 𝑖 sin 𝑛 𝜃, ∀𝑛 ∈ 𝑅 1 60 60 2 2 𝑧 2 = 4 (cos ( ) + 𝑖 sin ( )) 1 𝑧2 = 4 ( √3 2 1 +𝑖 ) 2 1 𝑧 2 = 2√3 + 2𝑖 Question #10 4𝜋 4𝜋 3 4 How do you find the fourth root of 16 (cos ( ) + 𝑖 sin ( ))? Answer: Step 1 De Moivre's formula tells us that: (cos 𝜃 + 𝑖 sin 𝜃)𝑛 = cos 𝑛 𝜃 + 𝑖 sin 𝑛 𝜃 So we find: 𝜋 𝜋 3 3 4 4𝜋 4𝜋 3 3 (2 (cos ( ) + 𝑖 sin ( ))) = 24 (cos ( ) + 𝑖 sin ( )) 4𝜋 4𝜋 3 3 = 16 (cos ( ) + 𝑖 sin ( )) 𝜋 𝜋 3 3 So 2 (cos ( ) + 𝑖 sin ( )) = 1 + √3𝑖 is a fourth root of 4𝜋 4𝜋 3 3 16 (cos ( ) + 𝑖 sin ( )) Is it the fourth root? No. Note that 4𝜋 3 is in Q3. So if we are using the normal range for arg(𝑧) ∈ (−𝜋, 𝜋] then the standard form of 4𝑝𝑖 3 is −2𝜋 3 and the principal fourth root is: 𝜋 𝜋 2 (cos (− ) + 𝑖 sin (− )) = √3 − 𝑖 6 6 which is in Q4. Here are the four fourth roots of 16 (cos ( 4𝑝𝑖 3 4𝜋 ) + 𝑖𝑠𝑖𝑛 ( )) 3 plotted in the Complex plane Question #11 How do you find all solutions to 𝑥 4 − 𝑖 = 0 ? Answer: Step 1 Given: 𝑥 4 = 𝑖 it is better to express the immaginary unit in exponential form 𝜋 𝑖 = 𝑒𝑖2 in such a way we can rewrite the equation like this 𝜋 𝑥 4 = 𝑒 𝑖( 2 +2𝑘𝜋) and as a consequence it must be 4 𝜋 𝑥 = √𝑒 𝑖( 2 +2𝑘𝜋) = 1 𝜋 𝑖( +2𝑘𝜋) 4 [𝑒 2 ] 𝜋 𝜋 = 𝑒 𝑖( 8 +𝑘 2 ) with 𝑘 = 0, 1, 2, 3 Question #12 How do I find a solution to 𝑒 2𝑥 = 8𝑒 𝑥 + 20 ? Answer: Step 1 𝑒 2𝑥 − 8𝑒 𝑥 − 20 = 0 By properties of exponents, we have: (𝑒 𝑥 )2 − 8𝑒 𝑥 − 20 = 0 We let 𝑡 = 𝑒 𝑥 𝑡 2 − 8𝑡 − 20 = 0 (𝑡 − 10)(𝑡 + 2) = 0 𝑡 = 10 and −2 𝑒 𝑥 = 10and 𝑒 𝑥 = −2 Take the natural logarithm of both sides. ln(𝑒 𝑥 ) = ln 1 0 and ln 𝑒 𝑥 = 𝑙𝑛 − 2 Use the rule log 𝑎𝑛 = 𝑛 log 𝑎: 𝑥 ln 𝑒 = ln 1 0 and 𝑥 ln 𝑒 = 𝑙𝑛 − 2 𝑙𝑛 and e are opposites, so equal 1 𝑥 = ln 1 0 and 𝑥 = 0 , since 𝑦 = ln 𝑥 has a domain of 𝑥 > 0 Question #13 How do I find the quotient Answer: −5+𝑖 −7+𝑖 ? Step 1 Multiply by the conjugate −5 + 𝑖 −7 − 𝑖 × −7 + 𝑖 −7 − 𝑖 35−2𝑖−(−1) 49−(−1) 36−2𝑖 50 = 18−𝑖 25 Question #14 How do I divide 6(cos ∘ 60 + 𝑖 sin 6 0∘ ) by 3(cos ∘ 90 + 𝑖 sin 9 0∘ )? Answer: Step 1 The easiest way to divide complex numbers is to first convert to polar form and then you may divide modulus separately and subtract the angles. Let 𝑧1 = 𝑟1 𝑐𝑖𝑠𝜃1 = 6𝑐𝑖𝑠60∘ = 6 cos 6 0∘ + 6𝑖 sin 6 0∘ = 6∠60∘ = 6∠ 𝑧2 = 𝑟2 𝑐𝑖𝑠𝜃2 = 3𝑐𝑖𝑠90∘ + 3𝑖 sin 9 0∘ = 3∠90∘ = 3∠ ∴ 𝑧1 𝑧2 = 𝑟1 𝑟2 𝜋 3 𝜋 2 𝑐𝑖𝑠(𝜃1 − 𝜃2 ) 6 = 𝑐𝑖𝑠(60∘ − 90∘ ) 3 = 2𝑐𝑖𝑠(−30∘ ) = 2∠ − 30∘ = 2(cos 3 0∘ − 𝑖 sin 3 0∘ ) = √3 − 𝑖 Question #15 Write the complex number 8 1+𝑖 in standard form? Answer: Step 1 In the division of complex number you need first to change the denominator into a pure real number. To do that you multiply and divide by the complex conjugate of the denominator, in this case, 1 − 𝑖 . So: 8 1 − 𝑖 8 − 8𝑖 8 8 × = = − 𝑖 = 4 − 4𝑖 1+𝑖 1−𝑖 1+1 2 2 Remember that: (1 + 𝑖)(1 − 𝑖) = 1 − 𝑖 + 𝑖 − 𝑖 2 but 𝑖 2 = √−12 = −1 giving: (1 + 𝑖)(1 − 𝑖) = 1 + 1 = 2 Question #16 Write the complex number 2+5𝑖 5+2𝑖 in standard form? Answer: Step 1 This is a division of complex numbers. We need first to transform the denominator into a real number; We do that multiplying and dividing by the complex conjugade of the denominator (5-2i) 2 + 5𝑖 5 − 2𝑖 × 5 + 2𝑖 5 − 2𝑖 = 10−4𝑖+25𝑖−10𝑖 2 25+4 But 𝑖 2 = −1 = 10 + 21𝑖 + 10 29 = = 20+21𝑖 29 20 29 + 21 29 𝑖 Which is in the form 𝑎 + 𝑏𝑖 Question #17 Write the complex number √3+𝑖 √3−𝑖 in standard form? Answer: Step 1 By rationalising the denominator, we get the standard form. √3 + 𝑖 √3 − 𝑖 Multiply and divide by (√3 + 𝑖) 2 ⇒ ⇒ (√3 + 𝑖) (√3 − 𝑖) × (√3 + 𝑖) (√3+𝑖) 2 3+1 ⇒( √3+𝑖 2 2 ) Question #18 How do you divide 6+5𝑖 ? 6−5𝑖 Answer: Step 1 6+5𝑖 6−5𝑖 (6+5𝑖)(6+5𝑖) = (6−5𝑖)(6+5𝑖) 36+60𝑖−25 = 36+25 11+60𝑖 = 61 Question #19 How do you divide 1 1+𝑖 ? Answer: Step 1 1 1−𝑖 1+𝑖 = = = (1+𝑖)(1−𝑖) 1−𝑖 1−𝑖 2 1−𝑖 1+1 = 1−𝑖 2 1 1 2 2 = − 𝑖 Question #20 What does (1+𝑖)×(6−2𝑖) 4𝑖 equal? Answer: Step 1 Given equation = (1+𝑖)(6−2𝑖) Expand the numerator = = = = 6−2𝑖+6𝑖−2𝑖 2 4𝑖 6+4𝑖−2(−1) 4𝑖 6+4𝑖+2 4𝑖 8+4𝑖 4𝑖 4 𝑖 Take 4 as common in numerator = = 4(2+𝑖) 4𝑖 2+𝑖 𝑖 multiply numerator and denominator with i = = 2𝑖+𝑖 2 𝑖2 2𝑖−1 −1 = 1 − 2𝑖