A complex number zz varies with tt such that z=2cos⁡t+i(3sin⁡t),where 0≤t<2π.z = 2\cos t + i(3\sin t),\quad \text{where } 0 \le t < 2\pi.

  1. (a)

    By taking x=Re⁡(z)x = \operatorname{Re}(z) and y=Im⁡(z)y = \operatorname{Im}(z), sketch on an Argand diagram the curve that shows the positions of the points representing the complex number zz. Find the cartesian equation of this curve.

  2. (b)

    Two complex numbers z1z_{1} and z2z_{2} for two distinct values of tt are such that ∣z1∣=∣z2∣|z_{1}| = |z_{2}| and 0<arg⁡(z1)<π20 < \arg(z_{1}) < \dfrac{\pi}{2}. By referring to the Argand diagram in part (a), find the possible values of arg⁡(z1+z2)\arg(z_{1} + z_{2}).

  3. (c)

    It is given further that z1z_{1} and z2z_{2} are roots to the quadratic equation z2+az+β=0z^{2} + a z + \beta = 0.

    Explain whether it is necessary for aa to be real.

  4. (d)

    Given that aa is not real and ∣z1∣=∣z2∣=262|z_{1}| = |z_{2}| = \dfrac{\sqrt{26}}{2}, find the values for aa and β\beta.

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Worked solution

8(a)

1. Express the real and imaginary parts of $z$ as parametric equations and eliminate $t$ to find the cartesian equation.

x=Re⁡(z)=2cos⁡t  ⟹  cos⁡t=x2x = \operatorname{Re}(z) = 2\cos t \implies \cos t = \frac{x}{2}
y=Im⁡(z)=3sin⁡t  ⟹  sin⁡t=y3y = \operatorname{Im}(z) = 3\sin t \implies \sin t = \frac{y}{3}

Using the identity cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1, we obtain the cartesian equation:
(x2)2+(y3)2=1\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1
x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1

2. Sketch the locus of $z$ on an Argand diagram based on the cartesian equation.

The equation x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1 describes an ellipse centered at the origin.
On an Argand diagram:

  • The intercepts on the real axis (Re⁡\operatorname{Re}) are at 22 and −2-2.
  • The intercepts on the imaginary axis (Im⁡\operatorname{Im}) are at 33 and −3-3.

As 0≤t<2π0 \le t < 2\pi, the locus traces out the entire ellipse.
(A continuous elliptical curve is drawn passing through (2,0)(2,0), (0,3)(0,3), (−2,0)(-2,0), and (0,−3)(0,-3) on the Argand plane.)

Answer

Cartesian equation: x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1
(Sketch of an ellipse centered at the origin on an Argand diagram, passing through 22 and −2-2 on the real axis, and 33 and −3-3 on the imaginary axis.)

8(b)

1. Identify the possible positions of $z_2$ using the symmetry of the ellipse in the Argand diagram.

From part (a), the curve is an ellipse centered at the origin, symmetric about the real and imaginary axes.
The condition ∣z1∣=∣z2∣|z_1| = |z_2| implies that z1z_1 and z2z_2 are equidistant from the origin, lying on the intersection of a circle and the ellipse.
Since 0<arg⁡(z1)<π20 < \arg(z_1) < \frac{\pi}{2}, z1z_1 lies strictly in the first quadrant. By symmetry, for a distinct point z2z_2, the possible positions are:

  1. Reflection across the real axis (4th quadrant): z2=zˉ1z_2 = \bar{z}_1
  2. Reflection across the imaginary axis (2nd quadrant): z2=−zˉ1z_2 = -\bar{z}_1
  3. Reflection through the origin (3rd quadrant): z2=−z1z_2 = -z_1

2. Determine $\arg(z_1 + z_2)$ for the reflection across the real axis.

If z2=zˉ1z_2 = \bar{z}_1:
z1+z2=z1+zˉ1=2Re⁡(z1)z_1 + z_2 = z_1 + \bar{z}_1 = 2\operatorname{Re}(z_1)
Since z1z_1 is in the first quadrant, Re⁡(z1)>0\operatorname{Re}(z_1) > 0.
The sum lies on the positive real axis, so arg⁡(z1+z2)=0\arg(z_1 + z_2) = 0.

3. Determine $\arg(z_1 + z_2)$ for the reflection across the imaginary axis and origin.

If z2=−zˉ1z_2 = -\bar{z}_1:
z1+z2=z1−zˉ1=2iIm⁡(z1)z_1 + z_2 = z_1 - \bar{z}_1 = 2i\operatorname{Im}(z_1)
Since z1z_1 is in the first quadrant, Im⁡(z1)>0\operatorname{Im}(z_1) > 0.
The sum lies on the positive imaginary axis, so arg⁡(z1+z2)=π2\arg(z_1 + z_2) = \frac{\pi}{2}.

If z2=−z1z_2 = -z_1:
z1+z2=0z_1 + z_2 = 0
The argument of 00 is undefined.

Therefore, the possible values for arg⁡(z1+z2)\arg(z_1 + z_2) are 00 and π2\frac{\pi}{2}.

Answer

00 or π2\frac{\pi}{2}

8(c)

1. Relate the coefficient $a$ to the sum of the roots.

By the sum of roots for the quadratic equation z2+az+β=0z^2 + az + \beta = 0, we have:
z1+z2=−az_1 + z_2 = -a

2. Use the possible values of $\arg(z_1+z_2)$ from part (b) to determine if $a$ must be real.

From part (b), one possible configuration (when z2=−zˉ1z_2 = -\bar{z}_1) gives arg⁡(z1+z2)=π2\arg(z_1 + z_2) = \frac{\pi}{2}, which means z1+z2z_1 + z_2 can be a purely imaginary number.
If z1+z2z_1 + z_2 is purely imaginary, then a=−(z1+z2)a = -(z_1 + z_2) is also purely imaginary (i.e. not real).
Hence, it is not necessary for aa to be real.

Answer

No, it is not necessary for aa to be real, because the sum of the roots z1+z2z_1 + z_2 can be purely imaginary.

8(d)

1. Use the given modulus to find the exact value of $z_1$.

Given ∣z1∣=262|z_1| = \frac{\sqrt{26}}{2}, we have ∣z1∣2=264=132|z_1|^2 = \frac{26}{4} = \frac{13}{2}.
Substituting z1=2cos⁡t1+i(3sin⁡t1)z_1 = 2\cos t_1 + i(3\sin t_1) into the modulus formula:
∣z1∣2=(2cos⁡t1)2+(3sin⁡t1)2=4cos⁡2t1+9sin⁡2t1|z_1|^2 = (2\cos t_1)^2 + (3\sin t_1)^2 = 4\cos^2 t_1 + 9\sin^2 t_1
4(1−sin⁡2t1)+9sin⁡2t1=1324(1-\sin^2 t_1) + 9\sin^2 t_1 = \frac{13}{2}
4+5sin⁡2t1=6.5  ⟹  5sin⁡2t1=2.5  ⟹  sin⁡2t1=124 + 5\sin^2 t_1 = 6.5 \implies 5\sin^2 t_1 = 2.5 \implies \sin^2 t_1 = \frac{1}{2}
Since 0<arg⁡(z1)<π20 < \arg(z_1) < \frac{\pi}{2} from part (b), z1z_1 must lie in the first quadrant, so sin⁡t1=12\sin t_1 = \frac{1}{\sqrt{2}} and cos⁡t1=12\cos t_1 = \frac{1}{\sqrt{2}}.
Thus, z1=2(12)+i(3(12))=2+322iz_1 = 2\left(\frac{1}{\sqrt{2}}\right) + i\left(3\left(\frac{1}{\sqrt{2}}\right)\right) = \sqrt{2} + \frac{3\sqrt{2}}{2}i.

2. Determine $z_2$ using the condition that $a$ is not real.

The equation z2+az+β=0z^2 + az + \beta = 0 gives the sum of roots as z1+z2=−az_1 + z_2 = -a.
Since aa is not real, z1+z2z_1 + z_2 must have a non-zero imaginary component (it cannot be purely real).
The other points on the curve with the same modulus of 262\frac{\sqrt{26}}{2} are z=±2±322iz = \pm \sqrt{2} \pm \frac{3\sqrt{2}}{2}i.
Testing the possible options for z2z_2:

  • If z2=2−322iz_2 = \sqrt{2} - \frac{3\sqrt{2}}{2}i (the conjugate), z1+z2=22z_1 + z_2 = 2\sqrt{2} (real).
  • If z2=−2−322iz_2 = -\sqrt{2} - \frac{3\sqrt{2}}{2}i (the negative), z1+z2=0z_1 + z_2 = 0 (real).
  • If z2=−2+322iz_2 = -\sqrt{2} + \frac{3\sqrt{2}}{2}i (reflection across yy-axis), z1+z2=32iz_1 + z_2 = 3\sqrt{2}i (imaginary, which satisfies aa is not real).
    Hence, z2z_2 must be −2+322i-\sqrt{2} + \frac{3\sqrt{2}}{2}i.

3. Calculate the values of $a$ and $\beta$.

From the sum of the roots:
−a=z1+z2=32i  ⟹  a=−32i-a = z_1 + z_2 = 3\sqrt{2}i \implies a = -3\sqrt{2}i

From the product of the roots, β=z1z2\beta = z_1 z_2:
β=(2+322i)(−2+322i)\beta = \left(\sqrt{2} + \frac{3\sqrt{2}}{2}i\right)\left(-\sqrt{2} + \frac{3\sqrt{2}}{2}i\right)
β=(322i)2−(2)2=−184−2=−4.5−2=−6.5=−132\beta = \left(\frac{3\sqrt{2}}{2}i\right)^2 - (\sqrt{2})^2 = -\frac{18}{4} - 2 = -4.5 - 2 = -6.5 = -\frac{13}{2}.

Answer

a=−32ia = -3\sqrt{2}i, β=−132\beta = -\frac{13}{2}

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