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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Additional resources for A First Course in Complex Analysis with Applications
The concept of a branch cut is important and will be examined in greater detail when we study functions in Chapters 2 and 4. 3 Polar Form of Complex Numbers 21 (iv ) The “cosine i sine” part of the polar form of a complex number is sometimes abbreviated cis. That is, z = r (cos θ + i sin θ) = r cis θ. This notation, used mainly in engineering, will not be used in this text. 3 Answers to selected odd-numbered problems begin on page ANS-2. In Problems 1–10, write the given complex number in polar form ﬁrst using an argument θ = Arg(z) and then using θ = Arg(z).
I127 − 5i9 + 2i−1 = quadrant. 33. 2i, is the farthest from the origin. 34. If 3i¯ z − 2z = 6, then z = . 35. If 2x − 3yi + 9 = −x + 2yi + 5i, then z = 36. If z = 5 √ , then Arg(z) = − 3+i . 37. If z = 0 is a real number, then z+z −1 is real. Other complex numbers z = x+iy for which z + z −1 is real are deﬁned by |z| = . 38. The position vector of length 10 passing through (1, −1) is the same as the complex number z = . √ quadrant. 39. The vector z = (2 + 2i)( 3 + i) lies in the 40. The boundary of the set S of complex numbers z satisfying both Im(y) > 0 .
EXAMPLE 1 Cube Roots of a Complex Number Find the three cube roots of z = i. Solution Keep in mind that we are basically solving the equation w3 = i. Now with r = 1, θ = arg(i) = π/2, a polar form of the given number is given by z = cos(π/2) + i sin(π/2). From (4), with n = 3, we then obtain wk = √ 3 1 cos π/2 + 2kπ 3 + i sin π/2 + 2kπ 3 , k = 0, 1, 2. 4 Powers and Roots 25 Hence the three roots are, k = 0, k = 1, k = 2, √ √ 4 = 2 and 3 27 = 3 are the principal square root of 4 and the principal cube root of 27, respectively.