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Complex Geometry TRANSFORMATIONS AND LOCI Transformations Last time, we looked at rotation of regular polygons, and how to combine them in weird and wonderful ways. Now weβre going to look at general complex functions, and how to transform them in other ways. The key points From last time, rotation around the origin is the same as multiplying by a complex number. E.g., multiplying by π is rotating clockwise by 90°. Dividing by a real number is the same as scaling from the origin, e.g. π§/π makes things π times bigger. Subtracting a complex number translates the graph, or equivalently adding a complex number translates it the βwrongβ way. Example Take a unitary regular complex polygon, and transform it so it is pointing down onto 3 + 2π. This is three steps: 1. Create the pentagon 2. Rotate the pentagon 90° Clockwise 3. Translate the pentagon so it points to 3 + 2π Example Take a unitary regular complex polygon, and transform it so it is pointing down onto 3 + 2π. 1. Create the pentagon π§5 β 1 = 0 Example Take a unitary regular complex polygon, and transform it so it is pointing down onto 3 + 2π. 1. Create the pentagon 2. Rotate the pentagon 90° Clockwise (π§ β π)5 β1 = 0 π§5π β 1 = 0 Example Take a unitary regular complex polygon, and transform it so it is pointing down onto 3 + 2π. 1. Create the pentagon 2. Rotate the pentagon 90° Clockwise 3. Translate the pentagon so it points to 3 + 2π Translate center to 3 + 3π: (π§ β (3 + 3π))5 π β 1 = 0 Complex Loci Complex numbers exist in two dimensions, meaning that when we place restrictions upon their location, we can draw curves or areas. There are two key ideas: writing an equation from a description, and working out the description from an equation. The first part is all about conics, the second is algebraic. Example 1 A point π = (π₯, π¦) is 3 units away from (3, β4). Describe π. This definition is clearly that of a circle. The general form of a circle is π₯ βπ 2 + π¦βπ The radius is 3, and (π, π) = (3, β4). Thus the final equation is π₯ β3 2+ π¦+4 2 = π2 2 = 32 Example 2 A point π = (π₯, π¦) is equidistant from (β1, 4) and (5, 2). This is trickier, as it doesnβt match any conic that I can think of. Instead, perhaps we should start with the direct translation? π₯, π¦ β β1, 4 = | π₯, π¦ β 5, 2 | π₯ + 1, π¦ β 4 = | π₯ β 5, π¦ β 2 | π₯+1 2 + π¦β4 2 = π₯+1 2 + π¦β4 2 = π₯β5 π₯β5 2 2 + π¦β2 + π¦β2 2 2 This is the perpendicular bisector of the line formed by the two points! Example 2 A point π = (π₯, π¦) is equidistant from (β1, 4) and (5, 2). π₯+1 2 + π¦β4 2 = π₯β5 2 + π¦β2 2 π₯+1 2 β π₯β5 2 = π¦β2 2 β π¦β4 2 π₯ 2 + 2π₯ + 1 β π₯ 2 + 10π₯ β 25 = π¦ 2 β 4π¦ + 4 β π¦ 2 + 8π¦ β 16 12π₯ β 24 = 4π¦ β 12 3π₯ β 6 = π¦ β 3 π¦ = 3π₯ β 3 Example 3 Describe the locus formed by the equation π§ β2 + π§+2 >5 where π§ is a complex variable. (π₯ + π¦π) β 2 + (π₯ + π¦π) + 2 > 5 π₯ β 2 + π¦π + π₯ + 2 + π¦π > 5 π₯β2 2 + π¦2 + π₯+2 2 + π¦2 > 5 π₯β2 2 + π¦2 > 5 β π₯+2 2 + π¦2 Example 3 π₯β2 2 + π¦2 > 5 β π₯β2 2 + π¦ 2 > 25 β 10 β π₯β2 2 + 25 < 10 π₯+2 2 + π¦2 π₯ 2 + 4π₯ + 4 β π₯ 2 + 4π₯ β 4 + 25 < 10 π₯+2 2 + π¦2 8π₯ + 25 < 10 π₯+2 2 + π¦2 π₯+2 2 8π₯ + 25 2 2 π₯+2 + π¦2 π₯+2 < 100 π₯ + 2 2 2 + π¦2 + π¦2 + π₯ + 2 2 + π¦2 Example 3 8π₯ + 25 2 < 100 π₯ + 2 2 + π¦2 64π₯ 2 + 400π₯ + 625 < 100 π₯ 2 + 4π₯ + 4 + π¦ 2 64π₯ 2 + 400π₯ + 625 < 100π₯ 2 + 400π₯ + 400 + 100π¦ 2 225 < 36π₯ 2 + 100π¦ 2 36π₯ 2 100π¦ 2 1< + 225 225 Example 3 36π₯ 2 100π¦ 2 1< + 225 225 1< 1< 62 π₯ 2 152 π₯2 5 2 2 + + 102 π¦ 2 152 π¦2 3 2 2 This is the area outside an ellipse, centered 5 on the origin, with semi-major axis length and semi-minor axis length 3 . 2 2 Do Now Any Questions? Delta Workbook Nothing this time. Workbook Pages 206-209, 227-232 This work is licensed under a Creative Commons AttributionNonCommercial-ShareAlike 4.0 International License. Aaron Stockdill 2016