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Consider the complex function f(z) = z 6 e 1 z (a) Find a Laurent Series expansion for f around z = 0. You should use the existing Taylor series for e z and use that as your starting point. (b) What kind of singularity is z = 0? Explain. (c) What is the residue of f at z = 0? Explain. (d) What is the value of I C z 6 e 1 z dz where C is positively oriented circle of radius 1, centered at the origin? (e) What is I C z 6 e 1 z dz where C is positively oriented circle of radius 1, centered at 1+i?

User Beartech
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(a) The Taylor series expansion of e^3z about z=0 is 1+3z+ 9/2 z^2 + 27/6z^3 +….

(b) The Laurent series expansion of f(z)= e^3z/z^3 about z=0 is 1/z^3 + 3/z^2 + 9/2z+ 27/6+….

(c) C−4​ = 1/72,C−3= 1/6,C−2​=− 1/2,C−1​=3,C0​=-20, C1=9.

(a) The Taylor series expansion of e^3z about z=0 can be found by taking derivatives of e^3z at z=0. The expansion starts with the constant term 1 and includes powers of z multiplied by the corresponding derivatives of e^3z.

(b) The Laurent series expansion of f(z)= e^3z/z^3 about z=0 involves expressing the function as a sum of terms with positive and negative powers of z. In this case, it begins with the term 1/z^3 and includes higher-order terms with coefficients determined by the derivatives of e^3z.

(c) From the Laurent series expansion in part (b), the values of the Laurent coefficients are as follows:

C−4​ = 1/72,C−3= 1/6,C−2​=− 1/2,C−1​=3,C0​=-20, C1=9.

These coefficients represent the coefficients of the terms with corresponding powers of z in the Laurent series expansion.

Complete ques:

Consider the complex function f(z) = e^3z/ z^3

(a) Find the Taylor series expansion of e 32 about the point z = 0.

(b) Find the Laurent series expansion of f(z) about the point z = 0.

(c) From the expansion that you found in part (b), state the values of the Laurent coefficients C-4, С-3, С-2, C-1, Co, and c₁.

User Celada
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