Video: Sketching Regions That the Complex Number Satisfies in the Complex Plane

The complex number 𝑧 satisfies the following conditions: 1. |𝑧| β©Ύ 2 |𝑧 + 12 βˆ’ 9𝑖|, 2. |𝑧 βˆ’ 2𝑖| β©Ύ |𝑧 + 6 + 4𝑖|, and 3. Im (𝑧) < 12. Represent the region on an Argand diagram.

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Video Transcript

The complex number 𝑧 satisfies the following conditions. The modulus of 𝑧 is greater than or equal to two times the modulus of 𝑧 plus 12 minus nine 𝑖. The modulus of 𝑧 minus two 𝑖 is greater than or equal to the modulus of 𝑧 plus six plus four 𝑖. And the imaginary part of 𝑧 is less than 12. Represent the region on an Argand diagram.

We’ll begin by considering the first region. We can find the centre and radius of the circle by substituting 𝑧 equals π‘₯ plus 𝑦𝑖 into our equation. Remember at the moment, we’re just finding the boundary for the region. We then square both sides of this equation. We can instantly replace two squared with four.

But for the other bit, we’re going to need to use the definition of the modulus. We know that the modulus of π‘₯ plus 𝑦𝑖 is equal to the square root of π‘₯ squared plus 𝑦 squared. So the left-hand side of our equation becomes π‘₯ squared plus 𝑦 squared. We then gather the real and imaginary parts on the right-hand side. And we get four times π‘₯ plus 12 all squared plus 𝑦 minus nine all squared. Expanding the parentheses and then simplifying, on the right-hand side, we get four π‘₯ squared plus 96π‘₯ plus four 𝑦 squared minus 72𝑦 plus 900. We subtract π‘₯ squared and 𝑦 squared from both sides of this equation. And then, we divide through by three.

Now we’re looking to find the Cartesian equation of a circle. So we’re going to complete the square in π‘₯ and 𝑦. For π‘₯, we get π‘₯ plus 16 all squared minus 256. And for 𝑦, we get 𝑦 minus 12 all squared minus 144. And we add 300 of course. Negative 256 minus 144 plus 300 is negative 100. So we add 100 to both sides. And we have the Cartesian equation of a circle. It has a centre at negative 16, 12 and a radius of 10 units.

The boundary for our first region is therefore this circle as shown. But how do we decide whether to shade inside or outside of this circle? Well, let’s choose a point that we know to be outside of the circle. Let’s choose the point whose Cartesian coordinates are one, zero. This is the complex number one. We’re going to substitute this into the first inequality and see if the statement makes sense.

This statement is the modulus of one is greater than or equal to two times the modulus of one plus 12 minus nine 𝑖. Or the modulus of one is greater than or equal to two times the modulus of 13 minus nine 𝑖. Well, the modulus of one is one. And the modulus of 13 minus nine 𝑖 is root 250. Well, it’s not true that one is greater than two root 250. So this statement is false. And that tells us we’re going to be interested in the inside of the circle. That’s the region that satisfies that first condition. We’ll fully shade a region when we’ve considered the other two situations.

Now for two, we know that the equation, the modulus of 𝑧 minus two 𝑖 equals the modulus of 𝑧 plus six plus four 𝑖, represents the perpendicular bisector of the line segment joining the point which represents two 𝑖 and negative six minus four 𝑖. That’s the line segment between zero, two and negative six, negative four. 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. Alternatively, in this example, we can do this by inspection. And we can see that the line passes through the point zero, negative four and negative four, zero. And in fact, it also passes through the centre of our circle.

Once again, we’ll substitute 𝑧 equals one into the inequality and see if the statement makes sense. The modulus of one minus two 𝑖 is the square root of five. And the modulus of one plus six plus four 𝑖 or the modulus of seven plus four 𝑖 is the square root of 65. Once again, we can see that it’s actually not true that the square root of five is greater than or equal to the square root of 65. And we can see that we’re interested in the other side of the line. And of course, remember, we drew a solid line because our inequality is a weak inequality.

Let’s now consider the third region. We’re told that the imaginary part of 𝑧 must be less than 12. The boundary of this region is the horizontal line passing through 12 on the imaginary axis. And because it’s a strict inequality, we draw a dash line. We’re interested in the region below this dash line. To satisfy all three regions in this question, we need the intersection of the regions. So we shade the overlap between the three regions. It’s the sector of the circle shown. And we are done. We have represented the region on an Argand diagram.

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