Solution to Dr. Z.'s Intro to Complex Variable Attendance Quiz for Lecture 23 1. Find the Taylor polynomial of degree 1 at z0=3i of the function f(z)=z^2+z Sol. to 1: FIRST WAY: T_1(f(z); z0)= f(z0)+ f'(z0)(z-z0) Here z0=3i. f(z)= z^2+z, hence f(z0)=(3i)^2+3i= -9+3i f'(z)=2z+1, hence f'(z0)=2(3i)+1= 1+ 6i Hence: Answer to 1: (-9+3i)+ (1+6i) (z-3i) THIS IS THE FINAL ANSWER! Check List: 1. The answer is a polynomial of degree ONE, NOT TWO. A few students, including a very good student forgot to plug in z=3i and gave the wrong answer (-9+3i)+ (2z+1) (z-3i). This is NONSENSE (in this context). 2. DO NOT ATTEMPT TO SIMPLIFY (-9+3i)+ (1+6i) (z-3i) further. This is the FINAL answer. The FORMAT of a Taylor polynomial (or series) around z=z0 is in terms of (z-z0) NOT z. WARNING: If in the final you would expand the above answer,expressing it in terms of z, you would get NO CREDIT. ---------------------------------------- SECOND WAY: Let w=z-3i, so z=3i+w f(z)=z^2+z=(3i+w)^2+ (3i+w)= (3i)^2+ 2(3i)(w)+w^2 + 3i+w= (-9+ 3i)+ (1+6i) w + w^2 Since we want the Taylor polynomial of degree ONE, we ignore the w^2 and get (-9+ 3i)+ (1+6i) w But w is an ABBREVIATION for z-3i so Answer to 1: (-9+3i)+ (1+6i) (z-3i) ------------------------------------- WATCH OUT: The answer (1+6i) (z-3i) + (-9+3i) (GIVEN BY THE BEST STUDENT in class) is not quite right, in the context of TAYLOR polynomials (or series) you have to stick to the right format, order matters, even though that 2z+1 is the same as 1+2z.