Complex numbers – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   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 \(a^2+b^2\).

   1. Add or subtract the real parts and the imaginary parts separately. \(z+w=4+2i\). \(2z-3w=(6-4i)-(3+12i)=\) \(3-16i\) So \(z+w=4+2i\) and \(2z-3w=3-16i\).
   2. Expand the brackets and use \(i^2=-1\), so \(-8i^2=+8\). \(zw=3+12i-2i-8i^2=3+10i+8=\) \(11+10i\)
   3. The conjugate of \(w\) is \(1-4i\) and \((1+4i)(1-4i)=1+16=17\). \(\dfrac{(3-2i)(1-4i)}{(1+4i)(1-4i)}=\dfrac{3-12i-2i+8i^2}{17}=\dfrac{-5-14i}{17}=\) \(-\dfrac5{17}-\dfrac{14}{17}i\) Check: \(w\times\left(-\tfrac5{17}-\tfrac{14}{17}i\right)=\tfrac{-5-14i-20i+56}{17}=3-2i\).
   4. The product of a number and its conjugate is always real: \(zz^*=|z|^2\). \(zz^*=(3-2i)(3+2i)=9+4=\) 13, so \(|z|=\) \(\sqrt{13}\)

2. Answer to Question 2

   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.

   1. Use the quadratic formula; the discriminant is negative. \(z=\dfrac{6\pm\sqrt{36-100}}2=\dfrac{6\pm8i}{2}=\) \(3\pm4i\) The roots form a conjugate pair.
   2. The discriminant is \(4-40=-36\), so \(\sqrt{-36}=6i\). \(z=\dfrac{-2\pm\sqrt{4-40}}2=\dfrac{-2\pm6i}2=\) \(-1\pm3i\) Check: sum \(-2\) and product \(1+9=10\).
   3. For real coefficients the conjugate is also a root; then use \(z^2-(\text{sum})z+\text{product}=0\). The other root is \(2-i\). Sum \(=4\), product \(=4+1=5\), so \(z^2-4z+5=0\). Check: \((2+i)^2-4(2+i)+5=3+4i-8-4i+5=0\).
   4. The roots are non-real when the discriminant is negative. Need \(k^2-36<0\), so \(-6\lt k<6\).

3. Answer to Question 3

   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 \(-\pi\lt \arg z\le\pi\).

   1. \(r=\sqrt2\), first quadrant, \(\theta=\tfrac\pi4\): \(\sqrt2\left(\cos\tfrac\pi4+i\sin\tfrac\pi4\right)\)
   2. \(r=2\), second quadrant, reference angle \(\tfrac\pi6\), \(\theta=\tfrac{5\pi}6\): \(2\left(\cos\tfrac{5\pi}6+i\sin\tfrac{5\pi}6\right)\)
   3. \(r=2\sqrt2\), third quadrant, reference angle \(\tfrac\pi4\), \(\theta=-\tfrac{3\pi}4\): \(2\sqrt2\left(\cos\left(-\tfrac{3\pi}4\right)+i\sin\left(-\tfrac{3\pi}4\right)\right)\)
   4. \(r=\sqrt{25+75}=10\), fourth quadrant, \(\theta=-\tfrac\pi3\): \(10\left(\cos\left(-\tfrac\pi3\right)+i\sin\left(-\tfrac\pi3\right)\right)\)

4. Answer to Question 4

   Method: in modulus-argument form, multiplying multiplies the moduli and adds the arguments; dividing divides the moduli and subtracts the arguments. The conjugate has the same modulus and the opposite argument.

   1. Multiply moduli, add arguments: \(|z_1z_2|=8\), \(\arg=\tfrac\pi3+\tfrac\pi6=\tfrac\pi2\). So \(z_1z_2=8(\cos\tfrac\pi2+i\sin\tfrac\pi2)=\) \(8i\)
   2. Divide moduli, subtract arguments: modulus \(\tfrac12\), argument \(\tfrac\pi3-\tfrac\pi6=\) \(\tfrac\pi6\)
   3. \(|z_1|^2=4\) and \(\arg=2\times\tfrac\pi3=\) \(\tfrac{2\pi}3\)
   4. \(\arg z_2^*=-\tfrac\pi6\), so \(\arg(z_1z_2^*)=\tfrac\pi3-\tfrac\pi6=\) \(\tfrac\pi6\)

   So working in modulus-argument form avoids expanding brackets altogether.

5. Answer to Question 5

   1. A circle, centre \((2,0)\), radius 3: \((x-2)^2+y^2=9\).
   2. The boundary and inside of a circle, centre \((-1,1)\), radius 2: \((x+1)^2+(y-1)^2\le4\).
   3. A half-line starting at the point \((1,0)\) (not included) at \(45^\circ\) above the real direction: \(y=x-1,\ x>1\).
   4. Perpendicular bisector of the points \(1\) and \(-3\): \((x-1)^2+y^2=(x+3)^2+y^2\Rightarrow-2x+1=6x+9\Rightarrow\) \(x=-1\).

6. Answer to Question 6

   Method: to find \(\sqrt{a+bi}\), write \((x+iy)^2=a+bi\) and compare real and imaginary parts: \(x^2-y^2=a\) and \(2xy=b\). Eliminate \(y\) to get a quadratic in \(x^2\); \(x\) must be real, so reject the negative value of \(x^2\).

   1. Let \((x+iy)^2=8-6i\): \(x^2-y^2=8\) and \(2xy=-6\). Then \(y=-3/x\) and \(x^4-8x^2-9=0\), so \((x^2-9)(x^2+1)=0\), \(x=\pm3\). Roots: \(\pm(3-i)\). Check: \((3-i)^2=9-6i-1=8-6i\).
   2. \(x^2-y^2=-15\), \(xy=4\): \(x^4+15x^2-16=0\), \((x^2+16)(x^2-1)=0\), \(x=\pm1\), \(y=\pm4\). Roots: \(\pm(1+4i)\). Check: \(1-16+8i\).
   3. \(x^2-y^2=3\), \(xy=2\): \(x^4-3x^2-4=0\), \((x^2-4)(x^2+1)=0\), so \(z=\pm(2+i)\).

7. Answer to Question 7

   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 \(a^2+b^2\).

   1. \(|Z|=\sqrt{9+16}=\) 5 ohms. \(I=\dfrac{20(3-4i)}{25}=\) \(2.4-3.2i\)
   2. \(|I|=\sqrt{5.76+10.24}=\) 4 A. \(\arg I=-\arctan\tfrac43=\) \(-0.927\) rad. The argument is negative, so the current lags the voltage.
   3. \(\dfrac1{Z}=\dfrac{3-4i}{25}\) and \(\dfrac1{Z_2}=\dfrac{6+8i}{100}\). Sum \(=\dfrac{12-16i+6+8i}{100}=\dfrac{18-8i}{100}\). So \(Z_T=\dfrac{100}{18-8i}=\dfrac{100(18+8i)}{388}=\dfrac{450+200i}{97}\approx\) \(4.64+2.06i\) ohms.

8. Answer to Question 8

   1. \(45\cos\tfrac\pi6=\tfrac{45\sqrt3}2\) and \(45\sin\tfrac\pi6=\tfrac{45}2\): \(z_A=\dfrac{45\sqrt3}2+\dfrac{45}2i\)
   2. Multiplying by a number of modulus 1 and argument \(\tfrac{2\pi}3\) rotates \(z_A\) anticlockwise through \(120^\circ\) about the origin. \(|z_B|=\) 45 and \(\arg z_B=\tfrac\pi6+\tfrac{2\pi}3=\) \(\tfrac{5\pi}6\).
   3. The tips are at arguments \(\tfrac\pi6,\tfrac{5\pi}6,\tfrac{3\pi}2\). \(z_A+z_B=45(\cos\tfrac\pi6+\cos\tfrac{5\pi}6)+45i(\sin\tfrac\pi6+\sin\tfrac{5\pi}6)=0+45i=45i\) and \(z_C=45(0-i)=-45i\). The sum is 0.
   4. The tips form an equilateral triangle inscribed in a circle of radius 45, so the side is \(45\sqrt3\). Check: \(|z_A-z_B|=|45\sqrt3|=\) \(45\sqrt3\approx77.9\) m.

9. Answer to Question 9

   Method: multiplying by \(i\) rotates a vector through \(90^\circ\) anticlockwise. To rotate about \(C\) instead of the origin, rotate the vector \(p-c\), then add \(c\) back.

   1. \(p-c=-3+2i-1-i=-4+i\). \(i(-4+i)=-4i-1\). So \(p^\prime=1+i-1-4i=\) \(-3i\). So the drone moves to \(-3i\), that is the point \((0,-3)\).
   2. The distance between two points is the modulus of the difference. \(p^\prime-p=-3i-(-3+2i)=3-5i\), so \(PP^\prime=\sqrt{9+25}=\) \(\sqrt{34}\approx5.83\) m. Check: the rotation keeps \(|p-c|=\sqrt{17}\), and \(PP'=\sqrt2\times\sqrt{17}=\sqrt{34}\).
   3. Points on the real axis have the form \(x+0i\); substitute into the equation of the circle. \(|x+0i-1-i|=5\) gives \((x-1)^2+1=25\), \((x-1)^2=24\), \(x=1\pm2\sqrt6\) (about \(5.90\) and \(-3.90\)). So the circle crosses the real axis at \(1+2\sqrt6\) and \(1-2\sqrt6\).

10. Answer to Question 10

    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 \(a^2+b^2\).

    1. \(G(100)=\dfrac1{1+i}=\dfrac{1-i}2\), \(|G|=\dfrac1{\sqrt2}\approx0.707\), \(\arg G=-\tfrac\pi4\approx-0.785\). \(G(300)=\dfrac1{1+3i}=\dfrac{1-3i}{10}\), \(|G|=\dfrac1{\sqrt{10}}\approx0.316\), \(\arg G=-\arctan3\approx-1.249\).
    2. \(|1+0.01i\omega|=\sqrt{1+10^{-4}\omega^2}\), and \(|1/a|=1/|a|\). The denominator grows with \(\omega\), so \(|G|\) falls: the filter weakens high frequencies.
    3. Need \(\sqrt{1+10^{-4}\omega^2}=2\), so \(10^{-4}\omega^2=3\), \(\omega=100\sqrt3\approx\) 173 rad s\(^{-1}\).

11. Answer to Question 11

    1. With \(z=x+iy\): \(x^2+y^2=(x-8)^2+(y-6)^2\). Expanding, \(0=-16x+64-12y+36\), so \(4x+3y=25\).
    2. Range: \(|z|\le10\). Substituting \(y=\dfrac{25-4x}3\): \(9x^2+(25-4x)^2=900\), \(25x^2-200x-275=0\), \(x^2-8x-11=0\), \(x=4\pm3\sqrt3\). Then \(y=3\mp4\sqrt3\). Points \((4+3\sqrt3,\,3-4\sqrt3)\) and \((4-3\sqrt3,\,3+4\sqrt3)\).
    3. The distance between them is \(\sqrt{(6\sqrt3)^2+(8\sqrt3)^2}=\sqrt{108+192}=\sqrt{300}=\) \(10\sqrt3\approx17.3\) km. (The line is 5 km from \(L_1\), so the chord is \(2\sqrt{100-25}\).)

12. Answer to Question 12

    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 \(a^2+b^2\).

    1. \(z_1=i\). \(z_2=i^2+i=\) \(-1+i\). \(z_3=(-1+i)^2+i=-2i+i=\) \(-i\). \(z_4=(-i)^2+i=-1+i\).
    2. \(z_4=z_2\), so the values repeat \(-1+i,-i,-1+i,-i,\dots\) for ever. Both have modulus at most \(\sqrt2<2\), so the sequence never escapes: \(c=i\) is inside.
    3. \(z_1=1+i\). \(z_2=(1+i)^2+1+i=2i+1+i=\) \(1+3i\). \(z_3=(1+3i)^2+1+i=1+6i-9+1+i=\) \(-7+7i\). \(|z_3|=7\sqrt2\approx9.9>2\), so \(c=1+i\) escapes.

13. Answer to Question 13

    1. \((-2)^3+6(4)+21(-2)+26=-8+24-42+26=\) 0.
    2. Divide by \((z+2)\): \(z^3+6z^2+21z+26=(z+2)(z^2+4z+13)\). Then \(z=\dfrac{-4\pm\sqrt{16-52}}2=\) \(-2\pm3i\).
    3. The real parts are \(-2,-2,-2\), all negative, so the bridge is stable.
    4. Points \((-2,0)\), \((-2,3)\) and \((-2,-3)\) lie on a vertical line. Their sum is \(-6\), which equals \(-b/a\) for the cubic.

14. Answer to Question 14

    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 \(-\pi\lt \arg z\le\pi\).

    1. \(\cos\tfrac\pi3=\tfrac12\) and \(\sin\tfrac\pi3=\tfrac{\sqrt3}2\): \(3+1+\sqrt3\,i=\) \(4+\sqrt3\,i\).
    2. \(R=\sqrt{16+3}=\sqrt{19}\approx\) 4.359; \(\phi=\arctan\dfrac{\sqrt3}4\approx\) 0.409 rad.
    3. \(h_1+h_2=\) \(\sqrt{19}\cos(\omega t+0.409)\). The greatest height is \(\sqrt{19}\approx\) 4.36 m.

15. Answer to Question 15

    1. \((2+i)^2=3+4i\). \((2+i)^3=(3+4i)(2+i)=2+11i\). \((2+i)^4=(3+4i)^2=-7+24i\). \(f=-7+24i-6(2+11i)+18(3+4i)-30(2+i)+25=(-7-12+54-60+25)+(24-66+72-30)i=0+0i=\) 0.
    2. The conjugate \(2-i\) is also a root. \((z-2-i)(z-2+i)=(z-2)^2+1=\) \(z^2-4z+5\).
    3. Divide: \(f(z)=(z^2-4z+5)(z^2-2z+5)\). The second factor has roots \(\dfrac{2\pm\sqrt{-16}}2=1\pm2i\). So \(z=2\pm i,\ 1\pm2i\).
    4. The points \((2,\pm1)\) and \((1,\pm2)\) form an isosceles trapezium with parallel sides of length 2 (at \(x=2\)) and 4 (at \(x=1\)) a distance 1 apart. Area \(=\tfrac12(2+4)\times1=\) 3.

16. Answer to Question 16

    1. \(|z_1+z_2|^2=(z_1+z_2)(z_1^*+z_2^*)=|z_1|^2+z_1z_2^*+z_2z_1^*+|z_2|^2\). Similarly \(|z_1-z_2|^2=|z_1|^2-z_1z_2^*-z_2z_1^*+|z_2|^2\). Adding, the cross terms cancel, giving \(2(|z_1|^2+|z_2|^2)\).
    2. \(|4-i|^2=17\), \(|2+3i|^2=13\), total 30. \(|z_1|^2=10\), \(|z_2|^2=5\), so \(2(10+5)=\) 30. They agree.
    3. The diagonals are \(z_1+z_2\) and \(z_1-z_2\). The sum of the squares of the diagonals equals the sum of the squares of all four sides, since the sides have lengths \(|z_1|,|z_2|,|z_1|,|z_2|\).
    4. Put \(z=x+iy\). \(z+\dfrac1z=x+iy+\dfrac{x-iy}{x^2+y^2}\), with imaginary part \(y\left(1-\dfrac1{x^2+y^2}\right)\). It is zero when \(y=0\) (real \(z\)) or \(x^2+y^2=1\) (that is \(|z|=1\)). Both cases are the only ones.

17. Answer to Question 17

    1. \(|0-3-4i|=\sqrt{9+16}=5\), so \(0\) is on the boundary. Equation: \((x-3)^2+(y-4)^2=25\).
    2. Put \(x=0\): \(9+(y-4)^2=25\), \((y-4)^2=16\), \(y=0\) or \(8\). The points are \(0\) and \(8i\).
    3. The centre is at distance 5 from the origin and the radius is 5. Greatest \(|z|=5+5=\) 10, at \(z=2(3+4i)=\) \(6+8i\). Least \(|z|=\) 0, at the origin.
    4. The tangent at the origin is perpendicular to \(3+4i\), so it has direction \(-4+3i\) (argument \(\pi-\arctan\tfrac34\)) or \(4-3i\) (argument \(-\arctan\tfrac34\)). The greatest \(\arg z\) approaches 2.498 and the least approaches \(-0.644\) (neither is reached exactly).
    5. The line \(x=3\) passes through the centre, so \(\tfrac12\) of the disc is taken: \(\tfrac{25\pi}2\approx39.3\).

18. Answer to Question 18

    1. The side from \(0\) to \(z_2\) has the same length as the side from \(0\) to \(z_1\) (the modulus is unchanged) and the angle between them is \(60^\circ\). An isosceles triangle with a \(60^\circ\) apex angle is equilateral.
    2. Let \(\omega=\cos\tfrac\pi3+i\sin\tfrac\pi3=\tfrac12+\tfrac{\sqrt3}2i\). Then \(\omega^2=-\tfrac12+\tfrac{\sqrt3}2i=\omega-1\). So \(z_1^2+z_2^2=z_1^2(1+\omega^2)=z_1^2\omega=z_1z_2\). Hence \(z_1^2+z_2^2=z_1z_2\).
    3. \(z_2=6\left(\tfrac12+\tfrac{\sqrt3}2i\right)=\) \(3+3\sqrt3\,i\). \(z_1^2=36\), \(z_2^2=9-27+18\sqrt3\,i=-18+18\sqrt3\,i\), \(z_1z_2=18+18\sqrt3\,i\). \(36-18+18\sqrt3\,i=18+18\sqrt3\,i\). Verified.
    4. Centroid \(=\tfrac13(9+3\sqrt3\,i)=3+\sqrt3\,i\), modulus \(\sqrt{9+3}=2\sqrt3\). Distance to \(z_1\): \(|3-\sqrt3\,i|=2\sqrt3\). Distance to \(z_2\): \(|2\sqrt3\,i|=2\sqrt3\). All equal \(2\sqrt3\).

19. Answer to Question 19

    1. The coefficients are real, so complex roots occur in conjugate pairs (conjugate root theorem).
    2. Product of roots \(=(3+2i)(3-2i)\alpha=13\alpha=13\), so \(\alpha=1\).
    3. Sum of roots \(=7=-p\), so \(p=-7\). Sum of pairs \(=13+6\alpha=19=q\). Product \(=13=-r\), so \(r=-13\). The cubic is \(z^3-7z^2+19z-13=0\).
    4. \((3+2i)^2=5+12i\), \((3+2i)^3=(5+12i)(3+2i)=-9+46i\). \(-9+46i-7(5+12i)+19(3+2i)-13=(-9-35+57-13)+(46-84+38)i=0\). Verified.
    5. \((z-1)(z^2-6z+13)\).

20. Answer to Question 20

    1. \(1+\cos\theta=2\cos^2\tfrac\theta2\) and \(\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2\). So \(1+z=2\cos\tfrac\theta2\left(\cos\tfrac\theta2+i\sin\tfrac\theta2\right)\). For \(-\pi\lt \theta\lt \pi\), \(\cos\tfrac\theta2>0\), so \(|1+z|=\) \(2\cos\tfrac\theta2\) and \(\arg(1+z)=\) \(\tfrac\theta2\).
    2. Similarly \(z-1=-2\sin^2\tfrac\theta2+2i\sin\tfrac\theta2\cos\tfrac\theta2=2i\sin\tfrac\theta2\left(\cos\tfrac\theta2+i\sin\tfrac\theta2\right)\). Dividing, \(\dfrac{z-1}{z+1}=\) \(i\tan\tfrac\theta2\), purely imaginary.
    3. \(\theta=\tfrac\pi2\): \(\dfrac{i-1}{i+1}=\dfrac{(i-1)(1-i)}{2}=\dfrac{i-i^2-1+i}{2}=\dfrac{2i}{2}=i\), and \(i\tan\tfrac\pi4=i\). Correct.
    4. \(\arg\dfrac{z-1}{z+1}=\pm\tfrac\pi2\) is the angle between the lines from \(z\) to \(1\) and from \(z\) to \(-1\). So the angle subtended by the diameter is \(90^\circ\), which is the angle in a semicircle theorem.