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A First Course in Complex Analysis with Applications by Dennis G. Zill

By Dennis G. Zill

Written for junior-level undergraduate scholars which are majoring in math, physics, laptop technology, and electric engineering.

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1) From (1), we can conclude that and ρn = r (2) cos nφ + i sin nφ = cos θ + i sin θ. 3. √ From (2), we define ρ = n r to be the unique positive nth root of the positive real number r. From (3), the definition of equality of two complex numbers implies that cos nφ = cos θ and sin nφ = sin θ. These equalities, in turn, indicate that the arguments θ and φ are related by nφ = θ + 2kπ, where k is an integer. Thus, φ= θ + 2kπ . n As k takes on the successive integer values k = 0, 1, 2, . . , √ n − 1 we obtain n distinct nth roots of z; these roots have the same modulus n r but different arguments.

Chapter 1 Review Quiz 45 Here the symbol A¯ means the conjugate of the matrix A, which is the matrix obtained by taking the conjugate of each entry of A. A¯T is then the transpose ¯ which is the matrix obtained by interchanging the rows with the columns. of A, The negative −A is the matrix formed by negating all the entries of A; the matrix A−1 is the multiplicative inverse of A. (a) Which of the following matrices are Hermitian, skew-Hermitian, or unitary?     A=     3i 10 −10 0 10 − 2i −4 + i −10 − 2i     4+i    −5i     B=     1 0 0 0 0 2+i √ 10 2+i √ 10 −2 + i √ 10 2−i √ 10            C=     2 1 + 7i 1 − 7i 4 −6 − 2i 1−i −6 + 2i     1+i    0 (b) What can be said about the entries on the main diagonal of a Hermitian matrix?

This entails assuming a particular solution of the form qp (t) = A sin γt + B cos γt, substituting this expression into the differential equation, simplifying, equating coefficients, and solving for the unknown coefficients A and B. It is left as an exercise to show that A = E0 X/(−γZ 2 ) and B = E0 R/(−γZ 2 ), where the quantities X = Lγ − 1/Cγ and Z = X 2 + R2 (14) are called, respectively, the reactance and impedance of the circuit. Thus, the steady-state solution or steady-state charge in the circuit is qp (t) = − E0 X E0 R sin γt − cos γt.

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