Complex Numbers (L8th)

Complex numbers up until Argand Diagram and Loci

Showing 26-35 of 35 problems
1993 Paper 1 Q5
D: 1500.0 B: 1516.0

If \(z=x+\mathrm{i}y\) where \(x\) and \(y\) are real, define \(\left|z\right|\) in terms of \(x\) and \(y\). Show, using your definition, that if \(z_{1},z_{2}\in\mathbb{C}\) then \(\left|z_{1}z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|.\) Explain, by means of a diagram, or otherwise, why \(\left|z_{1}+z_{2}\right|\leqslant\left|z_{1}\right|+\left|z_{2}\right|.\) Suppose that \(a_{j}\in\mathbb{C}\) and \(\left|a_{j}\right|\leqslant1\) for \(j=1,2,\ldots,n.\) Show that, if \(\left|z\right|\leqslant\frac{1}{2},\) then \[ \left|a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z\right|<1, \] and deduce that any root \(w\) of the equation \[ a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+1=0 \] must satisfy \(\left|x\right|>\frac{1}{2}.\)

1993 Paper 2 Q9
D: 1600.0 B: 1500.0

\textit{In this question, the argument of a complex number is chosen to satisfy \(0\leqslant\arg z<2\pi.\)} Let \(z\) be a complex number whose imaginary part is positive. What can you say about \(\arg z\)? The complex numbers \(z_{1},z_{2}\) and \(z_{3}\) all have positive imaginary part and \(\arg z_{1}<\arg z_{2}<\arg z_{3}.\) Draw a diagram that shows why \[ \arg z_{1}<\arg(z_{1}+z_{2}+z_{3})<\arg z_{3}. \] Prove that \(\arg(z_{1}z_{2}z_{3})\) is never equal to \(\arg(z_{1}+z_{2}+z_{3}).\)

1993 Paper 3 Q6
D: 1700.0 B: 1484.0

The point in the Argand diagram representing the complex number \(z\) lies on the circle with centre \(K\) and radius \(r\), where \(K\) represents the complex number \(k\). Show that $$ zz^* -kz^* -k^*z +kk^* -r^2 =0. $$ The points \(P\), \(Q_1\) and \(Q_2\) represent the complex numbers \(z\), \(w_1\) and \(w_2\) respectively. The point \(P\) lies on the circle with \(OA\) as diameter, where \(O\) and \(A\) represent \(0\) and \(2i\) respectively. Given that \(w_1=z/(z-1)\), find the equation of the locus \(L\) of \(Q_1\) in terms of \(w_1\) and describe the geometrical form of \(L\). Given that \(w_2=z^*\), show that the locus of \(Q_2\) is also \(L\). Determine the positions of \(P\) for which \(Q_1\) coincides with \(Q_2\).

1992 Paper 1 Q4
D: 1500.0 B: 1516.0

Sketch the following subsets of the complex plane using Argand diagrams. Give reasons for your answers.

  1. \(\{z:\mathrm{Re}((1+\mathrm{i})z)\geqslant0\}.\)
  2. \(\{z: |z^{2}| \leqslant2,\mathrm{Re}(z^{2})\geqslant0\}.\)
  3. \(\{z=z_{1}+z_{2}:\left|z_{1}\right|=2,\left|z_{2}\right|=1\}.\)

Show Solution
  1. Multiplication by \(1+i\) rotates by \(45^{\circ}\) anticlockwise
    TikZ diagram
  2. \(|z| \leq \sqrt{2}\), \(\textrm{Re}(z^2) \geq 0\) means \(\textrm{Re}{z} \geq \textrm{Im}{z}\)
    TikZ diagram
  3. These are all points within \(1\) unit from a circle radius \(2\) units.
    TikZ diagram
1992 Paper 2 Q10
D: 1600.0 B: 1529.8

Let \(\alpha\) be a fixed angle, \(0 < \alpha \leqslant\frac{1}{2}\pi.\) In each of the following cases, sketch the locus of \(z\) in the Argand diagram (the complex plane):

  1. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha,}\)
  2. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha-\pi,}\)
  3. \(|\dfrac{z-1}{z}|=1.\)
Let \(z_{1},z_{2},z_{3}\) and \(z_{4}\) be four points lying (in that order) on a circle in the Argand diagram. If \[ w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \] show, by considering \(\arg w\), that \(w\) is real.

Show Solution
TikZ diagram
TikZ diagram
TikZ diagram
TikZ diagram
\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real
1990 Paper 2 Q5
D: 1600.0 B: 1484.0

The distinct points \(L,M,P\) and \(Q\) of the Argand diagram lie on a circle \(S\) centred on the origin and the corresponding complex numbers are \(l,m,p\) and \(q\). By considering the perpendicular bisectors of the chords, or otherwise, prove that the chord \(LM\) is perpendicular to the chord \(PQ\) if and only if \(lm+pq=0.\) Let \(A_{1},A_{2}\) and \(A_{3}\) be three distinct points on \(S\). For any given point \(A_{1}'\) on \(S\), the points \(A_{2}',A_{3}'\) and \(A_{1}''\) are chosen on \(S\) such that \(A_{1}'A_{2}',A_{2}'A_{3}'\) and \(A_{3}'A_{1}''\) are perpendicular to \(A_{1}A_{2},A_{2}A_{3}\) and \(A_{3}A_{1},\) respectively. Show that for exactly two positions of \(A_{1}',\) the points \(A_{1}'\) and \(A_{1}''\) coincide. If, instead, \(A_{1},A_{2},A_{3}\) and \(A_{4}\) are four given distinct points on \(S\) and, for any given point \(A_{1}',\) the points \(A_{2}',A_{3}',A_{4}'\) and \(A_{1}''\) are chosen on \(S\) such that \(A_{1}'A_{2}',A_{2}'A_{3}',A_{3}'A_{4}'\) and \(A_{4}'A_{1}''\) are respectively perpendicular to \(A_{1}A_{2},A_{2}A_{3},A_{3}A_{4}\) and \(A_{4}A_{1},\) show that \(A_{1}'\) coincides with \(A_{1}''.\) Give the corresponding result for \(n\) distinct points on \(S\).

Show Solution
The perpendicular bisector of the chords runs through the origin, therefore \(LM\) is perpendicular to \(PQ\) if and only if \(\frac{l+m}{2}\) is perpendicular to \(\frac{p+q}{2}\), ie \begin{align*} && (l+m) &= it (p+q) \\ \Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\ \Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\ &&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\ &&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\ &&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right) \end{align*} Therefore as long as \(l+m, p+q \neq 0\) \(lm+pq = 0\) is equivalent to the chords being perpendicular. In the case where (say) \(l,m\) is a diameter, then the condition for the chords to be perpendicular is that \(p,q\) is also a diameter and at right angles, but clearly this is also equivalent to our condition. Suppose \(A_1, A_2, A_3\) are distinct points on \(S\), and \(A_1'\) is given and suppose \(a_i, a_i'\) are the corresponding complex numbers, then the conditions are: \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\ \\ \Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\ && a_3' &= -\frac{a_2a_3}{a_2'} \\ &&&= \frac{a_1'a_2a_3}{a_1a_2} \\ &&&= \frac{a_1'a_3}{a_1} \\ && a_1'' &= - \frac{a_3a_1}{a_3'} \\ &&&= \frac{a_3a_1a_1}{a_1'a_3} \\ &&&= \frac{a_1^2}{a_1'} \\ \Rightarrow && a_1'a_1'' &= a_1^2 \end{align*} Therefore \(a_1' = a_1''\) if \(a_1' = \pm a_1\) Suppose we have \(4\) points, then \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\ A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\ \\ \Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\ &&&= -\frac{a_1a_3a_4}{a_1'a_3} \\ &&&= -\frac{a_1a_4}{a_1'} \\ \Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\ &&&= \frac{a_4a_1a_1'}{a_1a_4} \\ &&&= a_1' \end{align*} So they coincide. For \(n\) points if there are an even number of points they coincide, an odd number and there are two points when they coincide.
1988 Paper 1 Q6
D: 1500.0 B: 1516.0

The complex numbers \(z_{1},z_{2},\ldots,z_{6}\) are represented by six distinct points \(P_{1},P_{2},\ldots,P_{6}\) in the Argand diagram. Express the following statements in terms of complex numbers:

  1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) and \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\,\);
  2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\).
If \((i)\) holds, show that \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\). Suppose that the statements \((i)\) and \((ii)\) both hold, and that \(z_{1}=0,\) \(z_{2}=1,\) \(z_{3}=z,\) \(z_{5}=\mathrm{i}\) and \(z_{6}=w.\) Determine the conditions which \(\mathrm{Re}(z)\) and \(\mathrm{Re}(w)\) must satisfy in order that \(P_{1}P_{2}P_{3}P_{4}P_{5}P_{6}\) should form a convex hexagon. Find the distance between \(P_{3}\) and \(P_{6}\) when \(\tan(\angle P_{3}P_{2}P_{6})=-2/3.\)

Show Solution
  1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) is equivalent to \(z_2 - z_1 = z_4 - z_5\). \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\) is equivalent to \(z_3-z_2 = z_5 - z_6\).
  2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) is equivalent to \(\frac{z_4 - z_2}{z_6-z_3} \in i\mathbb{R}\)
If \(z_2 - z_1 =z_4 - z_5\) and \(z_3-z_2 = z_5 - z_6\) then adding we get \(z_3 - z_1 = z_4 - z_6\) or \(z_4 - z_3 = z_6-z_1\), which is equivalent to \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\).
TikZ diagram
\(\textrm{Re}(z) > 1, \textrm{Re}(w) < 0, \textrm{Re}(z) +\textrm{Re}(w)=1\). (We only need one of the first two constraints, since the other is implied by the former). Since \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) we must have that \(\textrm{Im}(z) = \textrm{Im}(w)\). Combined with the vector logic we must have that \(\textrm{Im}(z) = \frac12\). Let \(z = a + \frac12i\) and \(w = (1-a) + \frac12i\). Since \(w - 1 = k(3-2i)(z-1)\) (the angle constraint) we must have that: \begin{align*} &&-a+\frac12i &= k(3-2i)((a-1) \frac12i) \\ &&&= k( 3 a - 2+(\frac72 - 2 a)i) \\ \Rightarrow && \frac{3a-2}{-a} &= \frac{\frac72-2a}{\frac12} \\ \Rightarrow && 3a-2&= 4a^2-7a \\ \Rightarrow && 0 &= 4a^2-10a+2 \\ \Rightarrow && a &= \frac{5 \pm \sqrt{17}}{4} \end{align*} Since \(a > 1, a = \frac{5 +\sqrt{17}}{4}\) and the distance is: \begin{align*} \left | z - w \right | &= | a+\frac12i - ((1-a) +\frac12i ) | \\ &= |2a-1| \\ &= \frac{5+\sqrt{17}}{2}-1 \\ &= \frac{3+\sqrt{17}}{2} \end{align*}
1988 Paper 2 Q3
D: 1600.0 B: 1530.2

The quadratic equation \(x^{2}+bx+c=0\), where \(b\) and \(c\) are real, has the properly that if \(k\) is a (possibly complex) root, then \(k^{-1}\) is a root. Determine carefully the restriction that this property places on \(b\) and \(c\). If, in addition to this property, the equation has the further property that if \(k\) is a root, then \(1-k\) is a root, find \(b\) and \(c\). Show that \[ x^{3}-\tfrac{3}{2}x^{2}-\tfrac{3}{2}x+1=0 \] is the only cubic equation of the form \(x^{3}+px^{2}+qx+r=0\), where \(p,q\) and \(r\) are real, which has both the above properties.

Show Solution
Suppose \(k\) is a root of our quadratic. There are two possibilities, if \(k^{-1} = k\) then we must have \(k^2 = 1\) so either \(\pm 1\) is a root or we must have \((x-k)(x-k^{-1}) = x^2+bx+c\). In the first case we can have: \begin{align*} x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\ x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\ x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\ \end{align*} In the other cases, \(c = 1\) and \(b = k^{-1}+k\). Therefore we must have \(c = 1\) and \(b\) can take any values. The statement "if \(k\) is a root then \(1-k\) is a root" implies these two roots are different, so we must have \(1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0\) so \(b = -1, c = 1\). Suppose \(x^3+px^2+qx+r = 0\) has the first property, then for any root \(k\) we must have: \(k^3 + pk^2 + qk + r = 0\) and \(1 + pk^{-1} + qk^{-2} + rk^{-3} = 0\) therefore \(x^3+px^2+qx+r\) and \(rx^3+qx^2+px+1 = 0\) must have identical roots (since \(x = \pm\) either wont work here since they imply having the roots \(1, 0\) or \(-1, 2, \frac12\) which is exactly our equation. Therefore \(r = 1, p = q\). Suppose \(x^3 + px^2 + px+1 = 0\) has the property that if \(k\) is a root \(1-k\) is a root, therefore: \begin{align*} 0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\ &= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\ &= -k^3+(3+p)k^2+(-3-3p)k+(2+2p) \end{align*} Since these roots must be the same as the original roots, we must have \(3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32\)
1988 Paper 2 Q4
D: 1600.0 B: 1516.0

The complex number \(w\) is such that \(w^{2}-2w\) is real.

  1. Sketch the locus of \(w\) in the Argand diagram.
  2. If \(w^{2}=x+\mathrm{i}y,\) describe fully and sketch the locus of points \((x,y)\) in the \(x\)-\(y\) plane.
The complex number \(t\) is such that \(t^{2}-2t\) is imaginary. If \(t^{2}=p+\mathrm{i}q\), sketch the locus of points \((p,q)\) in the \(p\)-\(q\) plane.

Show Solution
  1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
    TikZ diagram
  2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
    TikZ diagram
If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)
TikZ diagram
1988 Paper 3 Q3
D: 1700.0 B: 1500.0

Give a parametric form for the curve in the Argand diagram determined by \(\left|z-\mathrm{i}\right|=2.\) Let \(w=(z+\mathrm{i})/(z-\mathrm{i}).\) Find and sketch the locus, in the Argand diagram, of the point which represents the complex number \(w\) when \begin{questionparts} \item \(\left|z-\mathrm{i}\right|=2;\) \item \(z\) is real; \item \(z\) is imaginary. \end{questionpart}

Show Solution
There are many possible parametric forms, for example \(z = i + 2e^{it}, z = 2\ cos \theta + (1 + 2\sin \theta)i\) etc. It is a circle radius \(2\) about the point \(i\).
  1. \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{2i + 2e^{it}}{2e^{it}} \\ &= 2 + ie^{-it} \end{align*} This is obvious a circle radius \(1\) about the point \(2\).
    TikZ diagram
  2. If \(z\) is real, then \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(z+i)^2}{z^2+1} \\ &= \frac{z^2-1 + 2zi}{z^2+1} \end{align*} We can quickly notice this describes a circle radius \(1\) about \(0\). Alternatively, \(|z+i| = |z-i| \Rightarrow |\frac{z+i}{z-i}| = 1\) so we must be talking about points on the unit circle. Since this is a Mobius transform we know it maps lines and circles to lines and circles, therefore it must map to the unit circle;
  3. If \(z\) is purely imaginary, say \(it\) then: \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(it+i)(i-it)}{(-1+t)^2} \\ &= \frac{t^2-1}{(t-1)^2} \end{align*} Which is purely real, and can take all real values.
    TikZ diagram