Complex numbers – further maths test answers to download as PDF

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1. Answer to Question 1 (3 marks)

   Method: treat \(i\) like a letter and replace \(i^2\) by \(-1\). To divide, multiply the top and bottom by the conjugate of the denominator, so that the denominator becomes the real number \(a^2+b^2\). Mark scheme as shown in brackets.

   1. The conjugate of \(z_2\) is \(1+i\) and \((1-i)(1+i)=2\). \(\dfrac{(2+3i)(1+i)}{(1-i)(1+i)}=\dfrac{2+2i+3i-3}{2}=\dfrac{-1+5i}2\). [1 for conjugate, 1 for result] \(-\tfrac12+\tfrac52i\)
   2. Use \(|z_1z_2|=|z_1||z_2|\), which avoids multiplying out. \(|z_1||z_2|=\sqrt{13}\times\sqrt2=\) \(\sqrt{26}\) [1] Check: \(z_1z_2=5+i\) and \(|5+i|=\sqrt{26}\).

2. Answer to Question 2 (4 marks)

   Method: a polynomial with real coefficients has its non-real roots in conjugate pairs \(p\pm qi\). For a quadratic, use the formula with \(\sqrt{-k}=i\sqrt k\), or use the sum and product of the roots. Mark scheme as shown in brackets.

   1. Substitute \(z=1\). \(f(1)=1-3+7-5=0\). [1] So \((z-1)\) is a factor by the factor theorem.
   2. Divide by \((z-1)\) and solve the quadratic factor with the formula. \(f(z)=(z-1)(z^2-2z+5)\) [1]. \(z=\dfrac{2\pm\sqrt{-16}}2=\) \(1\pm2i\) (and 1) [1] So the roots are \(1\), \(1+2i\) and \(1-2i\).
   3. Points \((1,0)\), \((1,2)\), \((1,-2)\) on the line \(x=1\). [1] The two non-real roots are reflections of each other in the real axis.

3. Answer to Question 3 (4 marks)

   Method: for \(z=x+iy\), \(|z|=\sqrt{x^2+y^2}\). Plot the point first to find the quadrant, then find the reference angle \(\arctan\left|\dfrac yx\right|\) and adjust it so that \(-\pi\lt \arg z\le\pi\). Mark scheme as shown in brackets.

   1. \(u\) is in the second quadrant with reference angle \(\tfrac\pi3\); \(v\) is in the fourth quadrant with reference angle \(\tfrac\pi4\). \(|u|=\sqrt{9+27}=6\), \(\arg u=\tfrac{2\pi}3\) [1]; \(|v|=2\sqrt2\), \(\arg v=-\tfrac\pi4\) [1]
   2. Divide the moduli and subtract the arguments; \(\tfrac{11\pi}{12}\) lies in the required range. \(\left|\dfrac uv\right|=\dfrac6{2\sqrt2}=\) \(\dfrac{3\sqrt2}2\) [1]; \(\arg\dfrac uv=\tfrac{2\pi}3+\tfrac\pi4=\) \(\tfrac{11\pi}{12}\) [1]

4. Answer to Question 4 (6 marks)

   Method: \(|z-a|\) is the distance from \(z\) to the point \(a\). Points equidistant from two points lie on the perpendicular bisector; substitute \(z=x+iy\) and square to get Cartesian equations. Mark scheme as shown in brackets.

   1. \(|z-2|=|z-6i|\) [1]. \((x-2)^2+y^2=x^2+(y-6)^2\) [1] gives \(-4x+4=-12y+36\), so \(x-3y=-8\) [1]
   2. \((x-2)^2+y^2=25\) with \(x=3y-8\): \((3y-10)^2+y^2=25\) [1], \(10y^2-60y+75=0\), \(2y^2-12y+15=0\) [1], \(y=\dfrac{6\pm\sqrt6}2\), \(x=3y-8=\dfrac{2\pm3\sqrt6}2\). Points \(\left(\dfrac{2+3\sqrt6}2,\dfrac{6+\sqrt6}2\right)\) and \(\left(\dfrac{2-3\sqrt6}2,\dfrac{6-\sqrt6}2\right)\) [1]

   Check: the midpoint of \(AB\), \((1,3)\), satisfies \(1-9=-8\), as it should.

5. Answer to Question 5 (3 marks)

   Method: a polynomial with real coefficients has its non-real roots in conjugate pairs \(p\pm qi\). For a quadratic, use the formula with \(\sqrt{-k}=i\sqrt k\), or use the sum and product of the roots. Mark scheme as shown in brackets.

   1. \(2+i\) (conjugate) [1]
   2. Sum \(=4=-p\), so \(p=-4\) [1]. Product \(=(2-i)(2+i)=5=q\), so \(q=5\) [1]

   So the equation is \(z^2-4z+5=0\).

   Check: \((2-i)^2-4(2-i)+5=3-4i-8+4i+5=0\) ✓.

Total : 20 marks