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The complex number ๐ง satisfies the following conditions.
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The modulus of ๐ง is greater than or equal to two times the modulus of ๐ง plus 12 minus nine ๐.
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The modulus of ๐ง minus two ๐ is greater than or equal to the modulus of ๐ง plus six plus four ๐.
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And the imaginary part of ๐ง is less than 12.
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Represent the region on an Argand diagram.
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Weโll begin by considering the first region.
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We can find the centre and radius of the circle by substituting ๐ง equals ๐ฅ plus ๐ฆ๐ into our equation.
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Remember at the moment, weโre just finding the boundary for the region.
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We then square both sides of this equation.
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We can instantly replace two squared with four.
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But for the other bit, weโre going to need to use the definition of the modulus.
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We know that the modulus of ๐ฅ plus ๐ฆ๐ is equal to the square root of ๐ฅ squared plus ๐ฆ squared.
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So the left-hand side of our equation becomes ๐ฅ squared plus ๐ฆ squared.
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We then gather the real and imaginary parts on the right-hand side.
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And we get four times ๐ฅ plus 12 all squared plus ๐ฆ minus nine all squared.
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Expanding the parentheses and then simplifying, on the right-hand side, we get four ๐ฅ squared plus 96๐ฅ plus four ๐ฆ squared minus 72๐ฆ plus 900.
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We subtract ๐ฅ squared and ๐ฆ squared from both sides of this equation.
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And then, we divide through by three.
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Now weโre looking to find the Cartesian equation of a circle.
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So weโre going to complete the square in ๐ฅ and ๐ฆ.
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For ๐ฅ, we get ๐ฅ plus 16 all squared minus 256.
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And for ๐ฆ, we get ๐ฆ minus 12 all squared minus 144.
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And we add 300 of course.
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Negative 256 minus 144 plus 300 is negative 100.
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So we add 100 to both sides.
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And we have the Cartesian equation of a circle.
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It has a centre at negative 16, 12 and a radius of 10 units.
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The boundary for our first region is therefore this circle as shown.
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But how do we decide whether to shade inside or outside of this circle?
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Well, letโs choose a point that we know to be outside of the circle.
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Letโs choose the point whose Cartesian coordinates are one, zero.
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This is the complex number one.
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Weโre going to substitute this into the first inequality and see if the statement makes sense.
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This statement is the modulus of one is greater than or equal to two times the modulus of one plus 12 minus nine ๐.
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Or the modulus of one is greater than or equal to two times the modulus of 13 minus nine ๐.
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Well, the modulus of one is one.
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And the modulus of 13 minus nine ๐ is root 250.
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Well, itโs not true that one is greater than two root 250.
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So this statement is false.
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And that tells us weโre going to be interested in the inside of the circle.
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Thatโs the region that satisfies that first condition.
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Weโll fully shade a region when weโve considered the other two situations.
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Now for two, we know that the equation, the modulus of ๐ง minus two ๐ equals the modulus of ๐ง plus six plus four ๐, represents the perpendicular bisector of the line segment joining the point which represents two ๐ and negative six minus four ๐.
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Thatโs the line segment between zero, two and negative six, negative four.
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We could find the exact equation of the perpendicular bisector of this line segment by considering the gradient and midpoint of the line segment it bisects.
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Alternatively, in this example, we can do this by inspection.
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And we can see that the line passes through the point zero, negative four and negative four, zero.
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And in fact, it also passes through the centre of our circle.
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Once again, weโll substitute ๐ง equals one into the inequality and see if the statement makes sense.
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The modulus of one minus two ๐ is the square root of five.
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And the modulus of one plus six plus four ๐ or the modulus of seven plus four ๐ is the square root of 65.
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Once again, we can see that itโs actually not true that the square root of five is greater than or equal to the square root of 65.
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And we can see that weโre interested in the other side of the line.
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And of course, remember, we drew a solid line because our inequality is a weak inequality.
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Letโs now consider the third region.
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Weโre told that the imaginary part of ๐ง must be less than 12.
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The boundary of this region is the horizontal line passing through 12 on the imaginary axis.
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And because itโs a strict inequality, we draw a dash line.
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Weโre interested in the region below this dash line.
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To satisfy all three regions in this question, we need the intersection of the regions.
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So we shade the overlap between the three regions.
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Itโs the sector of the circle shown.
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And we are done.
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We have represented the region on an Argand diagram.