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Mathematics Class 10 Chapter 1: Complex Numbers

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This is the free study page for Mathematics Class 10, Chapter 1: Complex Numbers. It follows the Federal Board (FBISE) SSC-II syllabus. This chapter has 9 learning outcomes, and Parhayi.com has 259 MCQs, 49 short questions and 19 long questions for it. Read the topics and the key questions with their answers below, then tap Practice this chapter to try the questions yourself.

Mathematics grows step by step. Read the rule, study one worked example, then solve similar questions yourself and check every step.

What you will learn

Key questions and answers

These are the short questions of this chapter with their answers (49 questions).

Q1. Simplify \(\sqrt{-8}\,\sqrt{-2}\). (4 marks)

\(\sqrt{-8}=i\sqrt8,\ \sqrt{-2}=i\sqrt2\)write each root in terms of i first
\((i\sqrt8)(i\sqrt2)=i^2\sqrt{16}\)multiply
\(=-4\)\(i^2=-1\)

Q2. Express \(\dfrac{3+4i}{5-2i}\) in the form \(a+bi\). (4 marks)

\(\dfrac{3+4i}{5-2i}\times\dfrac{5+2i}{5+2i}\)multiply by the conjugate of the denominator
\(=\dfrac{15+6i+20i+8i^2}{25-4i^2}\)expand
\(=\dfrac{7+26i}{29}\)\(i^2=-1\)
\(=\dfrac{7}{29}+\dfrac{26}{29}i\)separate parts

Q3. Simplify \(\dfrac{1+i}{(2+i)(3-i)}\) and express the answer in the form \(a+bi\). (4 marks)

\((2+i)(3-i)=6-2i+3i-i^2=7+i\)multiply the denominator first
\(\dfrac{1+i}{7+i}\times\dfrac{7-i}{7-i}\)conjugate of 7+i
\(=\dfrac{7-i+7i-i^2}{49+1}=\dfrac{8+6i}{50}\)expand; \(i^2=-1\)
\(=\dfrac{4}{25}+\dfrac{3}{25}i\)reduce

Q4. Evaluate \(\dfrac{(1+i)^2}{6-2i}\) and express the result in the form \(a+bi\). (4 marks)

\((1+i)^2=1+2i+i^2=2i\)square the numerator
\(\dfrac{2i}{6-2i}=\dfrac{i}{3-i}\times\dfrac{3+i}{3+i}\)cancel 2, multiply by conjugate
\(=\dfrac{3i+i^2}{9+1}=\dfrac{-1+3i}{10}\)\(i^2=-1\)
\(=-\dfrac1{10}+\dfrac3{10}i\)final

Q5. Express \(\dfrac{2-3i}{(3+2i)^2}\) in the form \(a+bi\). (4 marks)

\((3+2i)^2=9+12i+4i^2=5+12i\)square the denominator
\(\dfrac{2-3i}{5+12i}\times\dfrac{5-12i}{5-12i}\)conjugate
\(=\dfrac{10-24i-15i+36i^2}{25+144}=\dfrac{-26-39i}{169}\)expand
\(=-\dfrac{2}{13}-\dfrac{3}{13}i\)reduce by 13

Q6. Find the additive inverse and the multiplicative inverse of \(z=3-4i\). (4 marks)

\(-z=-3+4i\)additive inverse: change both signs; check \(z+(-z)=0\)
\(\dfrac1z=\dfrac{1}{3-4i}\times\dfrac{3+4i}{3+4i}\)multiplicative inverse: rationalise
\(=\dfrac{3+4i}{9+16}=\dfrac{3}{25}+\dfrac{4}{25}i\)\(i^2=-1\)

Q7. Show that \(5-4i\) and \(\dfrac{5+4i}{41}\) are multiplicative inverses of each other. (4 marks)

\((5-4i)\cdot\dfrac{5+4i}{41}=\dfrac{(5)^2-(4i)^2}{41}\)\((a-b)(a+b)=a^2-b^2\)
\(=\dfrac{25+16}{41}=\dfrac{41}{41}=1\)\(i^2=-1\)
Product is 1, hence inverses.definition

Q8. Find the modulus of \(z=-2+2\sqrt3\,i\) and of \(\bar z\). (4 marks)

\(|z|=\sqrt{(-2)^2+(2\sqrt3)^2}=\sqrt{4+12}=4\)modulus formula
\(\bar z=-2-2\sqrt3\,i,\ |\bar z|=\sqrt{4+12}=4\)conjugate has the same modulus

Q9. If \(z_1=1-2i\) and \(z_2=2+i\), verify that \(\overline{z_1z_2}=\bar z_1\,\bar z_2\). (4 marks)

\(z_1z_2=2+i-4i-2i^2=4-3i\)multiply
\(\overline{z_1z_2}=4+3i\)LHS
\(\bar z_1\bar z_2=(1+2i)(2-i)=2-i+4i-2i^2=4+3i\)RHS
LHS = RHSverified

Q10. For \(z=2-5i\), show that \(\bar z\,z=|z|^2\). (4 marks)

\(\bar z z=(2+5i)(2-5i)=4-25i^2=29\)product with conjugate
\(|z|=\sqrt{4+25}=\sqrt{29}\Rightarrow|z|^2=29\)modulus squared
Both equal 29.proved

Q11. Separate \((\sqrt2-\sqrt3\,i)^2\) into real and imaginary parts. (4 marks)

\(=(\sqrt2)^2-2\sqrt2\sqrt3\,i+(\sqrt3\,i)^2\)\((a-b)^2\)
\(=2-2\sqrt6\,i-3=-1-2\sqrt6\,i\)\(i^2=-1\)
Real part −1, imaginary part \(-2\sqrt6\)answer

Q12. Find the real and imaginary parts of \(\dfrac{(2+3i)^2}{1-3i}\). (4 marks)

\((2+3i)^2=4+12i-9=-5+12i\)square
\(\dfrac{-5+12i}{1-3i}\times\dfrac{1+3i}{1+3i}=\dfrac{-5-15i+12i+36i^2}{10}\)conjugate
\(=\dfrac{-41-3i}{10}\)simplify
Re = \(-\tfrac{41}{10}\), Im = \(-\tfrac{3}{10}\)answer

Q13. Find the real and imaginary parts of \(\left(\dfrac{1+i}{1-i}\right)^{3}\). (4 marks)

\(\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}=\dfrac{(1+i)^2}{2}=\dfrac{2i}{2}=i\)simplify the base first
\(i^3=-i\)cube
Re = 0, Im = −1answer

Q14. Simplify \(i^{9}+i^{13}+i^{28}+(-i)^{21}\). (4 marks)

\(i^9=i,\ i^{13}=i,\ i^{28}=1\)remainders 1, 1, 0 on dividing by 4
\((-i)^{21}=-i^{21}=-i\)odd power keeps the minus
\(i+i+1-i=1+i\)add

Q15. Write \(4i^3+8i^6\) in the form \(a+bi\). (4 marks)

\(i^3=-i,\ i^6=(i^2)^3=-1\)powers of i
\(4(-i)+8(-1)=-8-4i\)substitute

Q16. Solve \(x^2+100=0\) and check one root. (4 marks)

\(x^2=-100\)shift
\(x=\pm\sqrt{-100}=\pm10i\)square root
\((10i)^2+100=100i^2+100=-100+100=0\)check

Q17. Determine whether \(1-i\) is a solution of \(x^2+2x+2=0\). (4 marks)

\((1-i)^2=1-2i+i^2=-2i\)square
\(-2i+2(1-i)+2=4-4i\ne0\)substitute
Not a solution.conclusion

Q18. Factorise \(x^2+25y^2\) over the complex numbers. (4 marks)

\(x^2+25y^2=x^2-25y^2i^2\)\(+1=-i^2\)
\(=(x)^2-(5yi)^2=(x-5yi)(x+5yi)\)difference of squares

Q19. Factorise \(3z^2+5\). (4 marks)

\(3z^2+5=(\sqrt3 z)^2-i^2(\sqrt5)^2\)\(5=-5i^2\)
\(=(\sqrt3z-i\sqrt5)(\sqrt3z+i\sqrt5)\)difference of squares

Q20. Solve the system \(z+w=3i,\ 2z+3w=2\). (4 marks)

\(2z+2w=6i\)multiply first equation by 2
\((2z+3w)-(2z+2w)=2-6i\Rightarrow w=2-6i\)subtract
\(z=3i-(2-6i)=-2+9i\)back substitute
\(2(-2+9i)+3(2-6i)=-4+18i+6-18i=2\)check

Q21. Solve \(z-4w=3i,\ 2z+3w=11-5i\). (4 marks)

\(2z-8w=6i\)×2
\(11w=11-11i\Rightarrow w=1-i\)subtract from second
\(z=3i+4(1-i)=4-i\)substitute

Q22. In a circuit \(E=IZ\). Find \(I\) if \(E=(70+220J)\) volts and \(Z=(16+8J)\) ohms. (4 marks)

\(I=\dfrac{70+220J}{16+8J}\times\dfrac{16-8J}{16-8J}\)divide using conjugate
\(=\dfrac{1120-560J+3520J-1760J^2}{256+64}\)expand
\(=\dfrac{2880+2960J}{320}=9+\dfrac{37}{4}J\) ampsimplify

Q23. Represent \(z_1=5+3i\), \(z_2=2-3i\), their sum and difference on the complex plane. (4 marks)

\(z_1+z_2=7+0i\)sum → (7,0)
\(z_1-z_2=3+6i\)difference → (3,6)
Plot (5,3), (2,−3), (7,0), (3,6); draw arrows from O; sum is the diagonal of the parallelogram on \(z_1,z_2\), difference the diagonal on \(z_1,-z_2\)parallelogram law

Q24. Represent the product of \(z_1=4+2i\) and \(z_2=-2+3i\) graphically. (4 marks)

\(z_1z_2=-8+12i-4i+6i^2=-14+8i\)multiply
Plot (4,2), (−2,3) and (−14,8) and draw arrows from O.geometric representation

Q25. If \(z=a+bi\), prove that \(z^2-(\bar z)^2\) is purely imaginary. (4 marks)

\(z^2=a^2-b^2+2abi\)square
\((\bar z)^2=a^2-b^2-2abi\)square of conjugate
\(z^2-(\bar z)^2=4abi\)subtract; real part 0

Q26. Find \(x\) and \(y\) if \((x+yi)(2+3i)=13\). (4 marks)

\(2x-3y+(3x+2y)i=13+0i\)expand
\(2x-3y=13,\ 3x+2y=0\)compare parts
\(x=2,\ y=-3\)solve: from second y=−3x/2; 2x+9x/2=13 ⇒ x=2

Q27. Find the value of \(m\) and \(n\) if \(3+3i=(m-2)i+n\). (4 marks)

\(n=3\)compare real parts
\(m-2=3\Rightarrow m=5\)compare imaginary parts

Q28. Find the multiplicative inverse of \(\dfrac{-i}{2-3i}\). (4 marks)

\(\text{inverse}=\dfrac{2-3i}{-i}\)flip
\(=\dfrac{2-3i}{-i}\times\dfrac{i}{i}=\dfrac{2i-3i^2}{1}=3+2i\)rationalise

Q29. If \(z=\dfrac{1-i}{10}\), evaluate \(\dfrac{1}{z-z^2}\). (4 marks)

\(z^2=\dfrac{-2i}{100}=-\dfrac{i}{50}\)square
\(z-z^2=\dfrac{5-5i+i}{50}=\dfrac{5-4i}{50}\)LCM 50
\(\dfrac{50}{5-4i}\times\dfrac{5+4i}{5+4i}=\dfrac{250+200i}{41}\)rationalise

Q30. Show that \(\left(\dfrac{1+i}{\sqrt2}\right)^2=i\) and hence find \(\left(\dfrac{1+i}{\sqrt2}\right)^{8}\). (4 marks)

\(\dfrac{(1+i)^2}{2}=\dfrac{2i}{2}=i\)square
\((i)^4=1\)eighth power = fourth power of the square

Q31. Write \(3\sqrt2-\sqrt{-7}\) in the form \(a+bi\) and find its conjugate. (4 marks)

\(\sqrt{-7}=\sqrt7\,i\)root of negative
\(3\sqrt2-\sqrt7\,i\)a+bi form
\(\text{conjugate}=3\sqrt2+\sqrt7\,i\)change sign of imaginary part

Q32. Simplify \(\dfrac{1+3i}{i\sqrt5}\). (4 marks)

\(\dfrac{1+3i}{\sqrt5 i}\times\dfrac{i}{i}=\dfrac{i+3i^2}{\sqrt5 i^2}=\dfrac{-3+i}{-\sqrt5}\)multiply by i
\(=\dfrac{3-i}{\sqrt5}=\dfrac{3\sqrt5}{5}-\dfrac{\sqrt5}{5}i\)rationalise

Q33. Find \(z\bar z\) for \(z=14-7i\) and \(z=-\dfrac12+i\). (4 marks)

\((14-7i)(14+7i)=196+49=245\)first
\(\left(-\tfrac12+i\right)\left(-\tfrac12-i\right)=\tfrac14+1=\tfrac54\)second

Q34. Find two complex numbers whose sum is \(6\) and whose product is \(13\). (4 marks)

\(x^2-6x+13=0\)form the quadratic from sum and product
\(x=\dfrac{6\pm\sqrt{36-52}}{2}=\dfrac{6\pm4i}{2}=3\pm2i\)quadratic formula
\((3+2i)+(3-2i)=6,\ (3+2i)(3-2i)=13\)check

Q35. If \(|z|=5\) and the real part of \(z\) is 3, find all possible \(z\). (4 marks)

\(9+b^2=25\Rightarrow b^2=16\)modulus squared
\(b=\pm4\Rightarrow z=3+4i\text{ or }3-4i\)two answers

Q36. Which of \(z_1=3-4i\) and \(z_2=-5+i\) is farther from the origin? Justify. (4 marks)

\(|z_1|=5\)3-4-5
\(|z_2|=\sqrt{26}\approx5.1\)modulus
\(z_2\) is farther (26 > 25).compare

Q37. Multiply \(z=2+3i\) by \(i\) and describe what happens to its point on the Argand plane. (4 marks)

\(iz=2i+3i^2=-3+2i\)multiply
(2,3) → (−3,2): same distance \(\sqrt{13}\) from O, turned 90° anticlockwise.geometric meaning

Q38. Prove that \(|z_1z_2|=|z_1||z_2|\) for \(z_1=3+4i\), \(z_2=1-i\). (4 marks)

\(z_1z_2=3-3i+4i-4i^2=7+i\)product
\(|7+i|=\sqrt{50}=5\sqrt2\)LHS
\(|z_1||z_2|=5\cdot\sqrt2=5\sqrt2\)RHS

Q39. Solve \(6y^2+36=0\). (4 marks)

\(y^2=-6\)divide by 6, shift
\(y=\pm\sqrt6\,i\)root

Q40. Find \(z\) if \(iz+2=5i\). (4 marks)

\(iz=5i-2\)shift 2
\(z=\dfrac{5i-2}{i}\times\dfrac{-i}{-i}=\dfrac{-5i^2+2i}{1}=5+2i\)divide by i
\(i(5+2i)+2=5i-2+2=5i\)check

Q41. Find the values of \(x\) and \(y\) if \(\dfrac{x}{1+i}+\dfrac{y}{1-i}=1\) (x, y real). (4 marks)

\(\dfrac{x(1-i)+y(1+i)}{2}=1\)LCM \((1+i)(1-i)=2\)
\((x+y)+(y-x)i=2\)compare parts
\(x+y=2,\ y-x=0\Rightarrow x=y=1\)solve

Q42. Give two examples of fields where complex numbers are used and state the formula \(E=IZ\) in words. (4 marks)

Electronics / electrical engineering, control systems, electromagnetism, computer graphics, mechanical engineering.fields
Voltage = current × impedance; in AC circuits all three can be complex.formula

Q43. Find \(|z_1+z_2|\), \(|z_1z_2|\) and \(\left|\dfrac{z_1}{z_2}\right|\) for \(z_1=1-2i,\ z_2=2+i\). (4 marks)

\(z_1+z_2=3-i\Rightarrow|3-i|=\sqrt{10}\)sum
\(|z_1z_2|=\sqrt5\cdot\sqrt5=5\)product property
\(|z_1/z_2|=\sqrt5/\sqrt5=1\)quotient property

Q44. Find the real and imaginary parts of \(\dfrac{(1+i)^2}{(1-2i)^2}\). (4 marks)

\((1+i)^2=2i,\ (1-2i)^2=-3-4i\)squares
\(\dfrac{2i}{-3-4i}\times\dfrac{-3+4i}{-3+4i}=\dfrac{-6i+8i^2}{25}=\dfrac{-8-6i}{25}\)rationalise
Re = \(-\tfrac{8}{25}\), Im = \(-\tfrac{6}{25}\)answer

Q45. Find the real and imaginary parts of \(\left(\dfrac{3+2i}{2-i}\right)^{2}\). (4 marks)

\(\dfrac{3+2i}{2-i}\times\dfrac{2+i}{2+i}=\dfrac{6+3i+4i+2i^2}{5}\)multiply by the conjugate of the denominator
\(=\frac{4}{5}+\frac{7}{5}i\)\(i^2=-1\), denominator \(4+1=5\)
\(\left(\frac{4}{5}+\frac{7}{5}i\right)^2=-\frac{33}{25}+\frac{56}{25}i\)square: \((a+bi)^2=a^2-b^2+2abi\)
\(\text{Re}=-\frac{33}{25},\ \text{Im}=\frac{56}{25}\)read off the parts

Q46. For \(z_1=2+i\) and \(z_2=3-4i\), verify that (i) \(\overline{z_1z_2}=\bar z_1\bar z_2\) and (ii) \(\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}\). (4 marks)

\(z_1z_2=(2+i)(3-4i)=10-5i\)multiply
\(\overline{z_1z_2}=10+5i\)change sign of imaginary part
\(\bar z_1\bar z_2=(2-i)(3+4i)=10+5i\)multiply conjugates
\(\text{(i) both sides are equal}\)verified
\(\left|\dfrac{z_1}{z_2}\right|=\dfrac{|2+i|}{|3-4i|}=\dfrac{\sqrt5}{5}\)\(\frac{z_1}{z_2}=\frac{2+11i}{25}\), modulus \(=\frac{\sqrt{125}}{25}=\frac{\sqrt5}{5}\)
\(\text{(ii) both sides are equal}\)verified

Q47. Find real numbers x and y if \(\dfrac{x}{1+i}+\dfrac{y}{1-i}=1+5i\). (4 marks)

\(\dfrac{x(1-i)}{2}+\dfrac{y(1+i)}{2}=1+5i\)multiply each term by the conjugate
\(\dfrac{(x+y)+(y-x)i}{2}=1+5i\)collect real and imaginary parts
\(x+y=2,\ y-x=10\)equate real and imaginary parts
\(x=-4,\ y=6\)add and subtract the equations
\(\dfrac{-4}{1+i}+\dfrac{6}{1-i}=-2+2i+3+3i=1+5i\)check

Q48. The points \(A(1+i)\), \(B(4+5i)\) and \(C(-3+4i)\) are given. Show that triangle ABC is an isosceles right-angled triangle. (4 marks)

\(AB=|(4+5i)-(1+i)|=|3+4i|=5\)distance formula
\(AC=|(-3+4i)-(1+i)|=|-4+3i|=5\)distance formula
\(BC=|(-3+4i)-(4+5i)|=|-7-i|=\sqrt{50}\)distance formula
\(AB^2+AC^2=25+25=50=BC^2\)Pythagoras converse
\(AB=AC\ \text{and}\ \angle A=90^\circ\)so triangle ABC is isosceles and right-angled

Q49. Simplify \((2+3i)^2-(2-3i)^2\). (3 marks)

\((2+3i)^2=4+12i+9i^2=-5+12i\)\((a+b)^2=a^2+2ab+b^2\) and \(i^2=-1\)
\((2-3i)^2=4-12i+9i^2=-5-12i\)same expansion
Difference \(=(-5+12i)-(-5-12i)=24i\)subtract

Long questions with answers

Q1. Solve the system of equations \((2+i)x+(1-i)y=3,\ (1+i)x-iy=2\). (8 marks)

\((2+i)x+(1-i)y=3\ \ldots(i)\)given
\((1+i)x-iy=2\ \ldots(ii)\)given
\(y=\dfrac{(1+i)x-2}{i}=\dfrac{[(1+i)x-2](-i)}{1}=(1-i)x+2i\)from (ii), make y the subject (divide by i = multiply by −i)
\((2+i)x+(1-i)[(1-i)x+2i]=3\)substitute in (i)
\((1-i)^2=-2i,\ (1-i)(2i)=2i-2i^2=2+2i\)simplify products
\((2+i)x-2ix+2+2i=3\Rightarrow(2-i)x=1-2i\)collect
\(x=\dfrac{1-2i}{2-i}\times\dfrac{2+i}{2+i}=\dfrac{2+i-4i-2i^2}{5}=\dfrac{4-3i}{5}\)divide
\(y=(1-i)\dfrac{4-3i}{5}+2i=\dfrac{4-3i-4i+3i^2}{5}+2i=\dfrac{1-7i}{5}+2i=\dfrac{1+3i}{5}\)back substitute

Q2. Let \(z_1=1-2i,\ z_2=2+i\). Show that (i) \(\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar z_1}{\bar z_2}\) and (ii) \(|z_1|=|-z_1|=|\bar z_1|=|-\bar z_1|\). (8 marks)

\(\dfrac{z_1}{z_2}=\dfrac{(1-2i)(2-i)}{5}=\dfrac{2-i-4i+2i^2}{5}=\dfrac{-5i}{5}=-i\)divide
\(\text{LHS}=\overline{-i}=i\)conjugate
\(\dfrac{\bar z_1}{\bar z_2}=\dfrac{(1+2i)(2+i)}{5}=\dfrac{2+i+4i+2i^2}{5}=i\)RHS; (i) proved
\(|1-2i|=|-1+2i|=|1+2i|=|-1-2i|=\sqrt5\)all four moduli; (ii) proved

Q3. Represent \(z_1=4+2i\) and \(z_2=2+3i\) and their sum and difference on the Argand plane, and explain the parallelogram law. (8 marks)

\(z_1+z_2=6+5i,\ z_1-z_2=2-i\)compute
Draw arrows from O to (4,2), (2,3), (6,5), (−2,−3) and (2,−1).plot
The sum is the diagonal of the parallelogram with sides \(z_1,z_2\); the difference is the diagonal of the parallelogram with sides \(z_1\) and \(-z_2\).parallelogram law
Alternative: the arrow from the tip of \(z_2\) to the tip of \(z_1\) also represents \(z_1-z_2\).note

Q4. Show that \(1+i\) and \(1-i\) are the roots of \(x^2-2x+2=0\), and hence factorise \(x^2-2x+2\) over \(\mathbb C\). (8 marks)

\((1+i)^2-2(1+i)+2=2i-2-2i+2=0\)check first root
\((1-i)^2-2(1-i)+2=-2i-2+2i+2=0\)check second root
\(x^2-2x+2=(x-1-i)(x-1+i)\)factorise
\((x-1)^2-i^2=x^2-2x+1+1\)verify

Practice MCQs

Choose the answer in your mind first. The answer is shown below each question.

1. The imaginary unit \(i\) is defined as:

  1. \(\sqrt{-1}\)
  2. \(\sqrt{1}\)
  3. \(-1\)
  4. \(1\)

Answer: a

\(i=\sqrt{-1}\), so \(i^2=-1\).

2. The multiplicative identity in \(\mathbb C\) is:

  1. \(1+0i\)
  2. \(1+1i\)
  3. \(0+0i\)
  4. \(0+1i\)

Answer: a

\(z\times(1+0i)=z\).

3. If \(I=(2-J)\) amp and \(Z=(2+J)\) ohm, then \(E\) equals:

  1. \(4-J\) volt
  2. \(5J\) volt
  3. \(5\) volt
  4. \(3\) volt

Answer: c

\((2-J)(2+J)=4+1=5\).

4. \((7-2i)+(-3-5i)\) equals:

  1. \(4-7i\)
  2. \(4-3i\)
  3. \(4+3i\)
  4. \(10+3i\)

Answer: a

Add real parts and imaginary parts separately: \(4-7i\).

5. The value of \(i^{25}+i^{-13}\) is:

  1. \(1\)
  2. \(i\)
  3. \(-i\)
  4. \(0\)

Answer: d

Reduce each power of i using \(i^4=1\). The result is \(0\).

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