# A better argument would be to argue that the reals

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A better argument would be to argue that the reals are closed under multiplication, examining the equation z 1/ n = x , x R and translating it to x n = z, and finally noting that r n must be real via closure of the group ( R , · ) , so z is real. 21
Joseph Heavner Honors Complex Analysis Assignment 2 January 25, 2015 1.5 Sets of Points in the Complex Plane 1.) Sketch the graph of | z - 4 + 3 i | = 5 . Figure 1: Circle with radius 5 centered at 4 - 3 i 3.) Sketch the graph of | z + 3 i | = 2 . Figure 2: Circle with radius 2 centered at - 3 i 5.) Sketch the graph of Re ( z ) = 5 . Figure 3: Vertical line at x = 5 1
7.) Sketch the graph of Im ( z + 3 i ) = 6 Unlike the previous problems, which were obvious enough as to call for no explanation, here we will write the equation in a more graph-able form. In particular, for z = a + bi I ( z + 3 i ) = 6 I ( a - bi + 3 i ) = 6 - b + 3 = 6 b = - 3 . Figure 4: Horizontal line at y = - 3 9.) Sketch the graph of | Re ( 1 + i z ) | = 3 | R ( 1 + i ¯ z ) | = 3 | R ( 1 + i ( x - iy )) | = 3 | 1 + y | = 3 ( 1 + y ) 2 = 9 y = 2 , - 4 . Figure 5: Horizontal lines at y = 2 and y = - 4 2
11.) Sketch the graph of Re ( z 2 ) = 1 R (( a + bi ) 2 ) = 1 R ( a 2 + 2 bi - b 2 ) = 1 a 2 - b 2 = 1 . Figure 6: Hyperbola x 2 - y 2 = 1 13.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. Re ( z ) < - 1 a.) yes b.) no c.) yes d.) no e.) yes Some explanation is that any neighborhood in the region R is entirely contained (i.e. any e -ball is entirely contained), therefore it is open. Also, not all of the boundary points of R are contained in R , therefore it is closed. Because any two points z 0 , z 1 ∈ R can be connected by a collection of connected lines, the set is connected; by definition, an open, connected set is a domain. The set is not bounded, however, as the modulus of a point z 0 ∈ R can become arbitrarily large (i.e. the set extends indefinitely towards x = - ). . Figure 7: Half-plane x < - 1 3
15.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. Im ( z ) > 3 a.) yes b.) no c.) yes d.) no e.) yes The justification here is identical to that in (13). . Figure 8: Half-plane y > 3 17.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. 2 < Re ( z - 1 ) < 4 To make this set easier to plot, we simplify the expression describing it a bit. 2 < R ( z - 1 ) < 4 2 < x - 1 < 4 3 < x < 5 a.) yes b.) no c.) yes d.) no e.) yes The justification here is identical to that in (13) and (15). . Figure 9: 3 < x < 5 4
19.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.
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