Questions and Answers #18 Complex Numbers

Questions and Answers Sheet 18 Complex Numbers Question #1 I want to find this limit for complex variable z lim (𝑧 + 1) sin ( 𝑧→−1 1 𝑧+1 ) In the real case I know sin(z) is bounded by -1,1, and the limit is 0. But in the complex case sin(𝑧) is not bounded. How can I find the limit or prove that it doesn't exist? Answer: One may first consider the sequence {𝑧𝑛 } of complex numbers such that 𝑧𝑛 + 1 = 1 𝑛𝜋 , 𝑛 → ∞, 𝑧𝑛 → −1, giving lim (𝑧𝑛 + 1) sin ( 𝑧𝑛→−1 1 𝑧𝑛+1 ) = lim 1 𝑛→∞ 𝑛𝜋 sin(𝑛𝜋) = 0. Then one may consider the sequence {𝑧𝑛 } of complex numbers such that 𝑧𝑛 + 1 = 1 𝑛𝜋𝑖 , 𝑛 → ∞, 𝑧𝑛 → −1, giving lim (𝑧𝑛 + 1) sin ( 𝑧𝑛→−1 1 𝑧𝑛+1 ) = lim 𝑛→∞ 𝑒 𝑛𝜋 −𝑒 −𝑛𝜋 2𝑛𝜋 =∞ Then considered limit lim (𝑧 + 1) sin ( 𝑧→−1 1 𝑧+1 ) does not exist. Question #2 I was trying to prove that where w is a complex number with absolute value smaller than 1 and 𝑟 ∈ 𝑅, 𝑟 < 1. I saw some solution online, which says we substitute 𝑤 = 𝑚𝑒 𝑖𝑥 so that the LHS becomes (𝑚 − 𝑟)2 . How is that true? We have (𝑤 − 𝑟)(𝑤 − 𝑟) = (𝑚𝑒 𝑖𝑥 − 𝑟)(𝑚𝑒 −𝑖𝑥 − 𝑟) = 𝑒 𝑖𝑥 𝑒 −𝑖𝑥 (𝑚 − 𝑟𝑒 −𝑖𝑥 )(𝑚 − 𝑟𝑒 𝑖𝑥 ) = (𝑚 − 𝑟𝑒 −𝑖𝑥 )(𝑚 − 𝑟𝑒 𝑖𝑥 ) Answer: The statement from the online solution looks wrong. 𝐿𝐻𝑆 = |𝑤|2 − (𝑤 + 𝑤 ̅)𝑟 + 𝑟 2 𝑅𝐻𝑆 = 1 − (𝑤 + 𝑤 ̅)𝑟 + |𝑤|2 𝑟 2 Now compare |𝑤|2 + 𝑟 2 with 1 + |𝑤|2 𝑟 2 Proof of original assertion. 1 ≥ 𝑟2 1 − |𝑤|2 ≥ 𝑟 2 (1 − |𝑤|2 ) = 𝑟 2 − 𝑟 2 |𝑤|2 1 + 𝑟 2 |𝑤|2 ≥ 𝑟 2 + |𝑤|2 Question #3 If z is complex variable, how to prove lim 𝑧 4 /|𝑧|2 = 0 𝑧→0 I have simplified to lim 𝑧 4 /𝑧𝑧̅ = lim 𝑧 3 /𝑧̅ 𝑧→0 𝑧→0 because |𝑧|2 = 𝑧𝑧̅ But how to progress? Answer: Let 𝑟, 𝜃 ∈ 𝑅 and write 𝑧 = 𝑟𝑒 𝑖𝜃 Then, 𝑧 → 0 iff 𝑟 → 0 Thus, we can substitute 𝑧 = 𝑟𝑒 𝑖𝜃 into the simplification that you obtained to get 3 lim(𝑟𝑒 𝑖𝜃 ) /𝑟𝑒 −𝑖𝜃 = lim 𝑟 2 𝑒 4𝑖𝜃 = 0 𝑟→0 𝑟→0 I prefer this method to the modulus method because it is applicable to many other similar cases. Question #4 Let 𝑎0 , 𝑎1 , … , 𝑎𝑛−1 , 𝑎𝑛 complex numbers with 𝑎𝑛 ≠ 0. If 𝑎𝑛−1 𝑓(𝑧) = |𝑎𝑛 + 𝑧 + ⋯+ 𝑎0 𝑧𝑛 | Exist lim 𝑓(𝑧) ? |𝑧|→∞ Answer: If we let 𝑧 → ∞, intuitively then we expect the terms containing z in the denominator to drop out leaving the limit as |𝑎𝑛 | . Let's see how our intuition holds out. We have ||a_𝑛 + 𝑎𝑛−1 𝑧 + ⋯+ 𝑎0 𝑧𝑛 |−|𝑎𝑛 || ≤ | 𝑎𝑛−1 𝑧 +⋯+ 𝑎0 𝑧𝑛 | The above inequality follows from the reverse triangl inequality. Now apply the regular triangle inequality | an −1 + z + a0 a || n −1 | + n z z +| a0 | zn If the proof is rather informal at this point you can just say that as 𝑧 → ∞ the right hand side above tends to 0 and the limit holds. Otherwise, it wouldn't be too difficult to follow up with a formal \epsilon-\delta proof. Question #5 Perform the indicated operation with complex numbers. (Simplify your answe completely.) (4 − 8𝑖)(9 + 7𝑖) Answer: Step 1 We have to perform the indication operation with complex numbers (4 − 8𝑖)(9 + 7𝑖) Step 2 (4 − 8𝑖)(9 + 7𝑖) = 4(9) + 4(7𝑖) − 9(8𝑖) − (8𝑖)(7𝑖) = 36 + 28𝑖 − 72𝑖 − 56𝑖 2 = 36 + 𝑖(28 − 72) − 56(−1) = 36 − 44𝑖 + 56 = 92 − 44𝑖 Hence, (4 − 8𝑖)(9 + 7𝑖) = 92 − 44𝑖 Question #6 Show that if 𝑧1 and are complex numbers then 𝑧̅1 𝑧2 = 𝑧 ̅̅̅ 1 𝑧 ̅̅̅ 2 Answer: Step 1 Given 𝑧1 and 𝑧2 are two complex numbers and we have to show, ( ̅1) = 𝑧 𝑧 ̅̅̅ 1 𝑧2 𝑧 ̅̅̅ 2 and 𝑧2 ≠ 0 Step 2 Since we know that, ∵ 𝑧1 𝑧2 = 𝑧1 ⋅ 𝑧2 𝑧 1 𝑧1 𝑧2 𝑧2 𝑧2 ∴ ( 1) = 𝑧1 ⋅ ( ) = 1 1 𝑧2 𝑧2 Now, we have to prove this ( ) = Let 𝑧2 = 𝑥2 + 𝑖𝑦2 1 ( 𝑥2 +𝑖𝑦2 (𝑥2 −𝑖𝑦2 ) )=( 2 +𝑖𝑦2 )(𝑥2 −𝑖𝑦2 ) (𝑥22 +𝑦22 ) ) (𝑥2 −𝑖𝑦2 ) 1 (𝑥22 +𝑦22 ) 𝑥22 +𝑦22 =( = (𝑥2 −𝑖𝑦2 ) ) = ((𝑥 ) ∴ 𝑥22 + 𝑦22 is a real number. = 1 𝑥22 +𝑦22 (𝑥2 +𝑖𝑦2 ) (𝑥22 +𝑦22 ) and 1 𝑧2 = 1 1 1 𝑧2 𝑧2 Hence ( ) = 1 1 𝑧2 𝑧2 ∴( )= = 𝑥2 +𝑖𝑦2 𝑧 𝑧1 𝑧2 𝑧2 ∴ ( 1) = 1 𝑥2 −𝑖𝑦2 = (𝑥2 +𝑖𝑦2 ) (𝑥22 +𝑦22 ) Hence Proved. Question #7 Subtract the following two complex numbers: (−5 + 3𝑖) − (4 + 7𝑖) . What do you notice? Answer: Step 1 Here 𝑧 = (−5 + 3𝑖) − (4 + 7𝑖) 𝑧 = (−5 − 4) + 3𝑖 − 7𝑖 = −9 + 𝑖[3 − 7] 𝑧 = −9 − 4𝑖 We notice here 𝑧1 = −5 + 3𝑖 which lies in 2 quadrant 𝑧2 = 4 + 7𝑖 lies in 1 quadrant But 𝑧1 − 𝑧2 lies in 3 quadrant Answer: (−9 − 4𝑖) Question #8 Use complex variables to find the sum of the Fourier Series: sin(𝜃) + sin(2𝜃) 22 + sin(3𝜃) 23 + where 𝜃 is a real variable. Answer: ∑+∞ 𝑛=1 sin(𝑛𝜃) 2𝑛 = Im ∑𝑛≥1 ( 𝑒 𝑖𝜃 2 𝑛 ) = Im ( 𝑒 𝑖𝜃 2−𝑒 𝑖𝜃 𝑒 𝑖𝜃 (2−𝑒 −𝑖𝜃 ) ) = Im ( 5−4 cos 𝜃 ) hence: ∑+∞ 𝑛=1 sin(𝑛𝜃) 2𝑛 = 2 sin 𝜃 5−4 cos 𝜃 . Question #9 Using complex variables to evaluate real trig integral 2𝜋 ∫0 cos(2𝜃) 1−2𝑎 cos(𝜃)+𝑎2 𝑑𝜃 , −1 < 𝑎 < 1 I did the substitution 𝑧 = 𝑒 𝑖𝜃 and then got cos(2𝜃) = 1/2(𝑧 2 + 1/𝑧 2 ) ,so 2𝜋 ∫ 0 cos(2𝜃) 1 1/2(𝑧 2 + 1/𝑧 2 ) 𝑑𝜃 = ∫ 𝑑𝑧. 1 − 2𝑎 cos(𝜃) + 𝑎2 𝑖 |𝑧|=1 1 − 𝑎𝑧 − 𝑎/𝑧 + 𝑎2 Did I make a mistake here? Answer: 𝑧 2 + 𝑧 −2 𝑧4 + 1 = (𝑎 − 𝑧)(𝑎 − 𝑧 −1 ) 𝑧(𝑎 − 𝑧)(𝑎𝑧 − 1) Should be pretty easy to compute the residues at 𝑧 = 0 and 𝑧 = 𝑎. (Just delete the corresponding factor from the denominator and plug in the value to compute the residues.) Question #10 What is the modulus of 6 + 7𝑖 ? Answer: The complex number's absolute value, or modulus, is as follows: |𝑎 + 𝑏𝑖| = √𝑎2 + 𝑏 2 Specifically, modulus = |6 + 7𝑖| = √62 + 72 = √36 + 49 = √85 Question #11 How to simplify 2+𝑖 1−𝑖 completely? Answer: A complex number 𝑎 + 𝑏𝑖 has the conjugate a-bi. A real number is the result of a complex number and its conjugate. This fact will be used to remove the complex number from the given expression's denominator. 2+𝑖 = (2+𝑖)(1+𝑖) (1−𝑖)(1+𝑖) 1−𝑖 2+2𝑖+𝑖–1 = 1+𝑖−𝑖+1 = 1+3𝑖 1 2 = + 2 3 2𝑖