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Unformatted text preview: Complex Analysis Fall 2007 Homework 3: Solutions 1.3.10. Claim: Using the branch of the square root function given in the problem, √ z 2 = z iff z = 0 or z = re iθ with r > 0 and 0 ≤ θ < π . Proof: ( ⇐ ) Suppose z = re iθ with r ≥ 0 and 0 ≤ θ < π . Then z 2 = r 2 e i 2 θ and 0 ≤ 2 θ < 2 π so that √ z 2 = ( r 2 ) 1 / 2 e i 2 θ/ 2 = re iθ = z. ( ⇒ ) Suppose z is not of the desired form. We will show that √ z 2 6 = z . In this case z = re iθ with r > 0 and π ≤ θ < 2 π . Then z 2 = r 2 e i 2 θ = r 2 e i (2 θ 2 π ) and since 0 ≤ 2 θ 2 π < 2 π we have √ z 2 = ( r 2 ) 1 / 2 e i (2 θ 2 π ) / 2 = re iθ e iπ = z. Since z 6 = 0, this is not the same as z . 1.3.34. If z = x + iy , it is shown on page 38 of our text that sin z = sin x cosh y + i cos x sinh y. We will use this below. Let A = { z  Re z  < π/ 2 } and B = C \ { z  Im z = 0 and  z  ≥ 1 } . We are asked to show that z 7→ sin z carries A onto B . Since sin( z ) = sin z and sin( z ) = sin( z ), it is sufficient to show that z 7→ sin z maps A = { z  ≤ Re z < π/ 2 and Im z ≥ } onto B = { z  Re z ≥ 0 and Im z ≥ } \ { z  Im z = 0 and Re z ≥ 1 } . We will argue geometrically. Fix x ∈ [0 , π/ 2) and consider the vertical ray defined by z = x + iy , y ≥ 0. Writing sin z = u + iv we find that the real and imaginary parts of the image of this ray under sin z satisfy u = sin x cosh y v = cos x sinh y. If x = 0 then we have, in fact, u = 0 and v = sinh y . As y ≥ 0, the points u + iv then trace out the nonnegative imaginary axis. If x 6 = 0 then we know that sin x 6 = 0 and so as y ≥ 0 varies, we find that the equations expressing u and v above give the standard parametrization of the portion of hyperbola...
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This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.
 Fall '11
 RyanDaileda

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