Questions and Answers #2 Complex Numbers

Questions and Answers Sheet 2 Complex Numbers Question #1 Having the following complex numbers A=5-2i and B=3+8i a) A+B b) B-A c) 𝐴 × 𝐵 d) What is the complex conjugate of A and the complex conjugate of B? 𝐵 e) (B divide by A) 𝐴 Answer: Step 1 a) Given complex numbers A=5-2i B=3+8i In addition of complex numbers real terms are added and imaginary terms are added together. (a+ib)+(c+id)=(a+c)+i(b+d) Step 2 so the sum of given numbers A+B=(5-2i)+(3+8i) =(5+3)+i(-2+8) =8+6i Step 3 b) In subtraction of complex numbers real terms are subtracted and imaginary terms are subtracted together. (a+ib)-(c+id)=(a-c)+i(b-d) Step 4 so the difference of the terms B-A=(3+8i)-(5-2i) =(3-5)+i(8+2) =-2+10i Step 5 c)The product of two complex numbers can be found by (𝑎 + 𝑖𝑏) × (𝑐 + 𝑖𝑑) = (𝑎𝑐 − 𝑏𝑑) + 𝑖(𝑎𝑑 + 𝑏𝑐) Step 6 Here the product is 𝐴 × 𝐵 = (5 − 2𝑖) × (3 + 8𝑖) = (5 × 3 − (−2) × 8) + 𝑖(8 × 5 + (−2) × 3) =31+34i Step 7 d)The complex conjugate of a+ib is a-ib a-ib is a+ib Step 8 The complex conjugate of A=5-2i ⇒ 𝐴 = 5 + 2𝑖 B=3+8i ⇒ 𝐵 = 3 − 8𝑖 Step 9 e) In complex numbers for division multiply the numerator and denominator by conjugate of denominator 𝑎+𝑖𝑏 = 𝑎+𝑖𝑏 × 𝑐−𝑖𝑑 𝑐+𝑖𝑑 𝑐+𝑖𝑑 𝑐−𝑖𝑑 (𝑎𝑐−𝑏𝑑)+𝑖(𝑎𝑑+𝑏𝑐) = 𝑐 2 +𝑑 2 Step 10 Here the division of B by A will give 𝐵 𝐴 = = (3+8𝑖)(5+2𝑖) = (5−2𝑖)(5+2𝑖) (15−16)+𝑖(40+6) −1 29 52 +22 46 +𝑖 29 Question #2 Add the following two complex numbers: (3 + 2i) + (5 - 3i). What do you notice? Answer: By Adding the following two complex numbers : (3+2i) + (5-3i) As it is known that (a+ib) + (c+id) will be (a+c) + i (b+d) so that (3+2i) + (5-3i) will be (3+5) + i (2-3) (8-i) will be answer So we notice that by adding two complex numbers real parts and imaginary parts added separately. Question #3 Find the quotient 4𝜋 𝑧1 = 50 (cos 𝑧2 = 5 (cos 3 𝜋 3 𝑧1 𝑧2 of the complex numbers + 𝑖 sin + 𝑖 sin 𝜋 3 4𝜋 3 ) and ) Leave the answer in polar form. Answer: If the complex numbers 𝑧1 and 𝑧2 S have the polar forms 𝑧1 = 𝑟1 (cos 𝜃1 + 𝑖 sin 𝜃1 ) 𝑎𝑛𝑑 𝑧2 = 𝑟2 (cos 𝜃2 + 𝑖 sin 𝜃2 ) Then the product and the quotient of two complex numbers is 𝑧1 𝑟 = 1 (cos(𝜃1 − 𝜃2 ) + 𝑖 sin(𝜃1 − 𝜃2 )), 𝑧2 ≠ 0 𝑧2 𝑟2 The given complex numbers are 𝑧1 = 50 (cos 4𝜋 3 + 𝑖 sin 4𝜋 3 ) and 𝑧2 = 5 (cos Substitute: 𝑟1 = 50, 𝑟2 = 5, 𝜃1 = 𝑧1 𝑧2 𝑧1 𝑧2 = = 𝑟1 𝑟2 50 5 4𝜋 3 , 𝜃2 = (cos(𝜃1 − 𝜃2 ) + 𝑖 sin(𝜃1 − 𝜃2 )) (cos ( 4𝜋 3 𝜋 4𝜋 3 3 − ) + 𝑖 sin ( 3𝜋 3𝜋 = 10 (cos ( ) + 𝑖 sin ( )) 3 3 𝜋 − )) 3 𝜋 3 𝜋 in 3 + 𝑖 sin 𝜋 3 ) = 10(cos(𝜋) + 𝑖 sin(𝜋)) The quotient of the complex number is 𝑧1 𝑧2 = 10(cos(𝜋) + 𝑖 sin(𝜋)) Question #4 Find T(t), N(t), 𝑎 𝑇 , and 𝑎𝑁 at the given time t for the curve r(t). 𝑟(𝑡) = 𝑡 2 𝑖 + 2𝑡𝑗, 𝑡 = 1 Answer: Consider 𝑟(𝑡) = 𝑡 2 𝑖 + 2𝑡𝑗 𝑇(𝑡) = 𝑟 ′(𝑡) ||𝑟 ′(𝑡)|| ′ (𝑡) Here, 𝑟 S = 𝑡 2 𝑖 + 2𝑡𝑗 Thus, 𝑇(𝑡) = = = = = 𝑟 ′(𝑡) ||𝑟 ′(𝑡)|| 2𝑡𝑖+2𝑗 ||2𝑡𝑖+2𝑗|| 2𝑡𝑖+2𝑗 √4𝑡 2 +4 2𝑡𝑖+2𝑗 2√𝑡 2 +1 𝑡𝑖+𝑗 √𝑡 2 +1 Therefore, 𝑇(1) = 𝑖+𝑗 √2 Substitute t=1 in T(t) Question #5 What is √𝑖? If 𝑖 = √−1 is √𝑖 imaginary? Is it used or considered often in mathematics? How is it notated? Answer: Let z=(a+bi) be a complex number which is a square root of i, that is 𝑖 = 𝑧 2 − (𝑎2 − 𝑏 2 ) + 2𝑎𝑏𝑖 Equating real and imaginary parts we have, 𝑎2 − 𝑏 2 = 0 2ab=1 The ywo real solutions to this pair of equations are 1 𝑎= √2 1 𝑏= √2 and 1 𝑎=− √2 1 𝑏=− . √2 The two square roots of i therefore are 1 ± (1 + 𝑖) √2 Question #6 How to combine complex powers? It is not posssible to combine complex powers in the usual way: (𝑥 𝑦 ) 𝑧 = 𝑥 𝑦𝑧 There was mention of multi-valued functions, is there some theory that makes this all work out correctly? Answer: For complex numbers the general way of defining 𝑥 𝑦 (with x nonzero) is as exp [(𝑙𝑦) where l is one of the logarithms of x. But x does not have a unique logarithm, it has many. If l is one logarithm of x then the general one is 𝑙 + 2𝑛𝜋 i with n an integer. Hence the general value of 𝑥 𝑦 is exp(𝑙𝑦 + 2𝜋𝑖𝑛𝑦) = exp(𝑙𝑦) exp(2𝜋𝑖𝑛𝑦) Before proceeding further note that if y is an integer then exp(2𝜋 𝑖𝑛 𝑦) − 1 so that 𝑥 𝑛 is uniquely determined. More generally if y is a rational with denominator d then exp (𝜋 iny) (and so 𝑥 𝑦 ) takes d different values. But if y is irrational, then 𝑥 𝑦 takes infinitely many variables. In the problem in hand 𝑥 𝑦 takes the values \𝑒𝑥[(𝑙𝑦) exp(2𝜋 𝑖𝑛 𝑦)) as above, and 𝑥 𝑦𝑧 the values exp(𝑙𝑦𝑧) exp(2𝜋 𝑖𝑛 𝑦𝑧) However in general each value of 𝑥 𝑦 gives rise to infinitely many values of (𝑥 𝑦 ) 𝑧 , namely exp((𝑙𝑦 + 2𝜋 𝑖𝑛 𝑦)𝑧 + 2𝜋 𝑖𝑚𝑧) so (𝑥 𝑦 ) 𝑧 has a ''doubly-infinite'' family of values. In general, there will be values of (𝑥 𝑦 )𝑧 which are not values of 𝑥 𝑦 so the ''paradox'' in the previous posting should be expected. Question #7 Why √−1 ⋅ −1 ≠ √−12 ? Answer: Isaac's answer is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides. √−1 = 𝑖 LHS-i, RHS=i 1 1 = 𝑖 √−1 1 𝐿𝐻𝑆 = = −𝑖, 𝑅𝐻𝑆 = −𝑖 𝑖 √1 √−1 = 1 𝑖 1 𝐿𝐻𝑆 = = −𝑖, 𝑅𝐻𝑆 = −𝑖 𝑖 √ 1 −1 = 1 𝑖 𝐿𝐻𝑆 = √−1 = 𝑖, 𝑅𝐻𝑆 = −𝑖 We can then see that the error must be assuming √1 √−1 =√ 1 −1 Question #8 Prove that 𝑛−1 𝑝𝑟𝑜𝑑𝑘=1 sin 𝑘𝜋 𝑛 = 𝑛−1 𝑛 2 Using 𝑛𝑡ℎ root of unity (𝑒 2𝑘𝑖𝜋 𝑛 𝑛 ) =1 Answer: Step 1 𝑘𝜋 𝑃 = ∏𝑛−1 𝑘=1 sin ( ) 𝑛 𝑖𝑘𝜋 𝑖𝑘𝜋 𝑛 −𝑒 − 𝑛 ) 1−𝑛 ∏𝑛−1 𝑘=1 (𝑒 = (2𝑖) −𝑖 2𝑖𝑘𝜋 𝑛(𝑛−1)𝜋 𝑛−1 ∏ (𝑒 𝑛 −1) 2 𝑛 𝑘=1 = (2𝑖)1−𝑛 = 𝑛−1 𝑘 (−2)1−𝑛 ∏𝑘=1 (𝜉 −1) 𝑒 𝑛−1 = 21−𝑛 ∏𝑘=1 (1−𝜉 𝑘) 2𝑖𝜋 where 𝜉 = 𝑒 𝑛 Now note that 𝑥 𝑛 − 1 = (𝑥 − 1) ∑𝑛−1 𝑘=0 and 𝑘 𝑥 𝑛 − 1 = ∏𝑛−1 𝑘=0 (𝑥 − 𝜉 ) Cancelling (x-1) we have 𝑛−1 𝑘 𝑘 ∏𝑛−1 𝑘=1 (𝑥 − 𝜉 ) = ∑𝑘=0 𝑥 Substituting x=1 we have 𝑘 ∏𝑛−1 𝑘=1 (1 − 𝜉 ) = 𝑛 1−𝑛 ∴ 𝑃 = 𝑛2 Step 2 In order to note that 𝑛−1 𝑛 𝑥 − 1 = ∏(𝑥 − 𝜉 𝑘 ), 𝑘=0 note that 1, 𝜉, ⋯ , 𝜉 𝑛−1 are roots of 𝑥 𝑛 − 1. Therefore by polynomial reminder theorem we have 𝑛−1 𝑛 𝑥 − 1 = 𝑄(𝑥) ∏(𝑥 − 𝜉 𝑘 ) 𝑘=0 Comparing degrees we find Q(x) has degree 0. Comparing highest coefficients we conclude 𝑄(𝑥) = 1 Step 3 We may instead use the identity |1 − e 2 ik n = k ), n k = 1, , n − 1 = 2sin( to establish immediately that 𝑛−1 1−𝑛 ∏𝑛−1 𝑘𝜋 𝑘=1 |1−𝑒 𝑃 = ∏ sin ( ) = 2 𝑛 2𝑖𝑘𝜋 𝑛 | , 𝑘=1 and = 21−𝑛 |∏𝑛−1 𝑘=1 (1 − 𝑒 2𝑖𝑘𝜋 𝑛 )| continue by applying the foregoing logic to the product to obtain 𝑃 = 𝑛21−𝑛 Question #9 Complex-number inequality |𝑧1 𝑧2 ⋯ 𝑧𝑚 − 1| ≤ 𝑒 𝑧1−1|+⋯+|𝑧𝑚−1|−1 Can anybody tell me how to prove the following inequality? Answer: The inequality in question bounds how far you can get from 1 by multiplying several complex numbers that may individually not be far from 1. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction. Lemma: Suppose |𝑧1 − 1| = 𝛼1 𝑎𝑛𝑑 |𝑧2 − 1| = 𝛼2 . 𝑇ℎ𝑒𝑛 |𝑧1 𝑧2 − 1| ≤ (1 + 𝛼1 )(1 + 𝛼2 ) − 1 Proof: We know 𝛼1 𝛼2 = |𝑧1 𝑧2 − 𝑧1 − 𝑧2 + 1|. By triangle inequality on the three points 𝑧1 𝑧2 , 𝑧1 + 𝑧2 − 1 , and 1, we have |𝑧1 𝑧2 − 1| ≤ |𝑧1 𝑧2 − 𝑧1 − 𝑧2 + 1| + |𝑧1 + 𝑧2 − 2| ≤ 𝛼1 𝛼2 + 𝛼1 + 𝛼2 = (1 + 𝛼1 )(1 + 𝛼2 ) − 1 Now, for several numbers, |𝑧1 𝑧2 ⋯ 𝑧𝑚 − 1| ≤ (1 + 𝛼1 )(1 + 𝛼2,⋯,𝑚 ) − 1 ≤ (1 + 𝛼1 )(1 + 𝛼2 )(1 + 𝛼3,⋯,𝑚 ) − 1 ⋮ ≤ (1 + 𝛼1 )(1 + 𝛼2 ) ⋯ (1 + 𝛼𝑚 ) − 1 where 𝑎2,⋯,𝑚 , for example, is my hopefully transparent abuse of notation to denote |𝑧2 ⋯ 𝑧𝑚 − 1|. Finally, since 1 + 𝑥 ≤ 𝑒 𝑥 for real x, the desired inequality follows. Question #10 Argument of the summ of two complex numbers Let r, s be positive real numbers and 𝜃, 𝜙 real numbers with |𝜃 − 𝜙| < . Then an argument of 𝑟𝑒 𝑖𝜃 + 𝑠𝑒 𝑖𝜙 lies between 𝜃 and 𝜙 Can someone give a short, clean proof of the statement above that doesn't rely on geometric intuition? Answer: You want to show that if you write 𝑟𝑒 𝑖𝜃 + 𝑠𝑒 𝑖𝜃 as 𝑡𝑒 𝑖𝜒 , then 𝜒 lies between 𝜃 and 𝜙 . Without loss of generality, assume 𝜃 ≤ 𝜙. Factoring out 𝑒 𝑖𝜃 you get 𝑟𝑒 𝑖𝜃 + 𝑠𝑒 𝑖𝜙 = 𝑒 𝑖𝜃 (𝑟 + 𝑠𝑒 𝑖(𝜙−𝜃) ). Since multiplying by 𝑒 𝑖𝜃 is just a rotation by an angle 𝜃 , it is enough to consider the case where 𝜃 = 0 and 0 < 𝜙 < 𝜋. In that case, you have 𝑟 + 𝑠𝑒 𝑖𝜙 = 𝑟 + 𝑠(cos(𝜙) + 𝑖 sin(𝜙)), and you want to express it in the form 𝑡(cos(𝜒) + 𝑖 sin(𝜒)) Looking at real and complex parts, you see that 𝑡 sin(𝜒) = 𝑠 sin(𝜙), and 𝑟 + 𝑠 cos(𝜙) = 𝑡 cos(𝜒) 𝜋 Assume first that 0 < 𝜙 < . Then 𝜒 must also lie in the first quadrant, since we need both sin(𝜒) and 2 cos(𝜒) to be positive (since t, s sin (𝜙) , and 𝑟 + 𝑠 cos(𝜙) are all positive). If, on the other hand, \frac{\pi}{2}<\phi<\pi , then \cos(\phi) is negative. If 𝑟 + 𝑠 cos(𝜙) > 0, then we need cos(𝜒) > 0 , so 𝜒 is in the first quadrant and automatically smaller than 𝜙 and we are done. If 𝑟 + 𝑠 cos(𝜙) is negative then we need 𝜒 in the second quadrant. 𝜋 𝑠 In the former case, 0 < 𝜒, 𝜙 < , then from sin(𝜒) = sin(𝜙) and since sin(𝑥) is increasing on 0 ≤ 𝑥 ≤ 𝜋 2 2 , then 0 < 𝜒 < 𝜙 if and only if 𝑠 𝑡 𝑡 < 1 , if and only if s 𝑠 2 , since all of r, s, and cos(𝜙) are positive. Since s and t are both positive, s 1 , if and only if s>t . Here, since cos(𝜃) < 0 , then you have again 2 𝑡 2 = ||𝑟 + 𝑠𝑒 𝑖𝜙 || = 𝑠 2 + 𝑟(𝑟 + 2𝑠 cos(𝜙)) < 𝑠 2 (since 𝑟 + 𝑠 cos(𝜙) < 0 in this situation), so you get t0 . First, we'll prove that 𝐴 ⊂ 𝐶 − {0} then we'll show that 𝐶 − {0} ⊂ 𝐴 Step 2 Let's show that 𝐴 ⊂ 𝐶 − {0}. This amounts to showing that 0 does not belong to A. Given any w,z such 1 1 𝑧 𝑧 that 𝑤 = exp ( ) and 0<|z| 0 . Thus, 1 Now, let's show that any non-zero complex number x can be written as 𝑥 = exp ( ) with 0<|z|b a0 or i<0 , but this isn't true.