Locus intersections and extreme values | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

At an intersection, both boundary equations are true

Worked example

Intersect a vertical line and a circle

Find the points satisfying $\operatorname{Re}z=1$ and $|z-(3+2i)|=3$ . Write $z=x+iy$ .

x=1

(x-3)^2+(y-2)^2=9

4+(y-2)^2=9

Answer

z=1+(2\pm\sqrt5)i

A sketch shows whether there should be zero, one or two intersections and which one is named in the question.

Extreme modulus means nearest or farthest from the origin

Since $|z|$ is the distance from O, draw a line from O through the centre of a circular region. If the full disc $|z-c|\le r$ does not contain O, then

\min|z|=|c|-r

\max|z|=|c|+r

If the disc contains O, the minimum is 0. For a partial disc or a combined region, the nearest or farthest point must also satisfy every other condition; check boundary intersections as additional candidates.

Extreme argument means a limiting ray from the origin

Greatest argument: upper tangent from O

For a circle that does not contain the origin, the extreme argument occurs where a ray from the origin is tangent to the circle.

When O lies outside a circle, rotate a ray from the positive real axis until it just touches the circle. At the touching point T, the radius CT is perpendicular to the tangent OT, so triangle OCT is right-angled.

Worked example

Find the limiting arguments of a disc

For $|z-(4+3i)|\le2$ , the centre C is 5 units from O and has argument $\tan^{-1}(3/4)$ . In the tangent triangle,

\sin(\angle COT)=\frac{CT}{OC}=\frac25

\theta_c=\tan^{-1}\left(\frac34\right)

\phi=\sin^{-1}\left(\frac25\right)

Answer

\max\arg z=\theta_c+\phi

\min\arg z=\theta_c-\phi

Your turnIndependent locus-value check [5 marks]

For $|z-(5+2i)|\le2$ , find the greatest possible value of $|z|$ and the greatest possible value of $\arg z$ .

Show worked answer

|5+2i|=\sqrt{29}

\theta_c=\tan^{-1}\left(\frac25\right)

\phi=\sin^{-1}\left(\frac{2}{\sqrt{29}}\right)

Answer

\max|z|=\sqrt{29}+2

\max\arg z=\theta_c+\phi

Examiner note

A tangent answer needs the centre, radius, right angle and correct angle addition or subtraction. If another condition cuts the circle, confirm that the tangent point is still inside the allowed region.

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.5)

At an intersection, both boundary equations are true

Worked example

Intersect a vertical line and a circle

Find the points satisfying $\operatorname{Re}z=1$ and $|z-(3+2i)|=3$ . Write $z=x+iy$ .

x=1

(x-3)^2+(y-2)^2=9

4+(y-2)^2=9

Answer

z=1+(2\pm\sqrt5)i

A sketch shows whether there should be zero, one or two intersections and which one is named in the question.

Extreme modulus means nearest or farthest from the origin

Since $|z|$ is the distance from O, draw a line from O through the centre of a circular region. If the full disc $|z-c|\le r$ does not contain O, then

\min|z|=|c|-r

\max|z|=|c|+r

Extreme argument means a limiting ray from the origin

Greatest argument: upper tangent from O

For a circle that does not contain the origin, the extreme argument occurs where a ray from the origin is tangent to the circle.

Worked example

Find the limiting arguments of a disc

For $|z-(4+3i)|\le2$ , the centre C is 5 units from O and has argument $\tan^{-1}(3/4)$ . In the tangent triangle,

\sin(\angle COT)=\frac{CT}{OC}=\frac25

\theta_c=\tan^{-1}\left(\frac34\right)

\phi=\sin^{-1}\left(\frac25\right)

Answer

\max\arg z=\theta_c+\phi

\min\arg z=\theta_c-\phi

Your turnIndependent locus-value check [5 marks]

For $|z-(5+2i)|\le2$ , find the greatest possible value of $|z|$ and the greatest possible value of $\arg z$ .

Show worked answer

|5+2i|=\sqrt{29}

\theta_c=\tan^{-1}\left(\frac25\right)

\phi=\sin^{-1}\left(\frac{2}{\sqrt{29}}\right)

Answer

\max|z|=\sqrt{29}+2

\max\arg z=\theta_c+\phi

Examiner note

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.5)