Hard Further Maths Complex numbers: arithmetic Questions

Challenging, exam-style Further Maths Complex numbers: arithmetic questions with worked solutions. Stretch yourself on the hardest complex-numbers, cubic-equations, conjugate-root-theorem, forming-equations problems.

complex-numberscubic-equationsconjugate-root-theoremforming-equationsconjugate-pairsdivision
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Given that x=−3+ix=-3+i is a root of the equation x4+5x3+2x2−22x−20=0x^{4}+5x^{3}+2x^{2}-22x-20=0, which of the following gives the remaining roots?
Show worked solution

Worked solution

  1. Write down what is given

    p(x)=x4+5x3+2x2−22x−20,x=−3+i is a rootp\left(x\right)=x^{4}+5x^{3}+2x^{2}-22x-20,\qquad x=-3+i\ \text{is a root}

    The coefficients of pp are real, which is the key to the whole question.

  2. Apply the conjugate root theorem

    x=−3+i is a root⇒x=−3−i is a rootx=-3+i\ \text{is a root}\Rightarrow x=-3-i\ \text{is a root}

    Non-real roots of a real polynomial always come in conjugate pairs.

  3. Form the quadratic factor from the conjugate pair

    (x−(−3+i))(x−(−3−i))=x2+6x+10\left(x-\left(-3+i\right)\right)\left(x-\left(-3-i\right)\right)=x^{2}+6x+10

    The pair multiplies out to a quadratic with real coefficients.

  4. Check the sum and product of that pair

    α+β=−6,αβ=10\alpha+\beta=-6,\quad\alpha\beta=10

    These reproduce the coefficients of the quadratic factor.

  5. Divide p(x)p\left(x\right) by the quadratic factor

    x4+5x3+2x2−22x−20=(x2+6x+10)(x2−x−2)x^{4}+5x^{3}+2x^{2}-22x-20=\left(x^{2}+6x+10\right)\left(x^{2}-x-2\right)

    Algebraic division leaves the remaining factor.

  6. Verify the factorisation by expanding

    (x2+6x+10)(x2−x−2)=x4+5x3+2x2−22x−20\left(x^{2}+6x+10\right)\left(x^{2}-x-2\right)=x^{4}+5x^{3}+2x^{2}-22x-20

    Expanding the product returns the original polynomial.

  7. Factorise the remaining quadratic

    x2−x−2=x2−x−2x^{2}-x-2=x^{2}-x-2

    The remaining quadratic has real roots.

  8. Solve the remaining quadratic

    x=−1 or x=2x=-1\text{ or }x=2

    These are the two real roots.

  9. Check the root x=−3−ix=-3-i in p(x)p\left(x\right)

    p(−3−i)=0p\left(-3-i\right)=0

    Substituting the root into the polynomial gives zero.

  10. Check the root x=−1x=-1 in p(x)p\left(x\right)

    p(−1)=0p\left(-1\right)=0

    Substituting the root into the polynomial gives zero.

  11. Check the root x=2x=2 in p(x)p\left(x\right)

    p(2)=0p\left(2\right)=0

    Substituting the root into the polynomial gives zero.

  12. Recall the defining property of ii

    i2=−1i^{2}=-1

    Every simplification of a complex product rests on this single fact.

  13. Recall the definition of a complex number

    z=a+bi,a,b∈Rz=a+bi,\quad a,b\in\mathbb{R}

    aa is the real part and bb is the imaginary part of zz.

  14. Recall the complex conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    The conjugate is obtained by reversing the sign of the imaginary part.

  15. Recall that a number times its conjugate is real

    (a+bi)(a−bi)=a2+b2\left(a+bi\right)\left(a-bi\right)=a^{2}+b^{2}

    This is why multiplying by the conjugate clears ii from a denominator.

  16. State the remaining roots

    x≠−3+i:x=−3−i or x=−1 or x=2x\neq-3+i:\quad x=-3-i\text{ or }x=-1\text{ or }x=2

    These are the other 33 roots of the equation.

Answer
x=−3−i or x=−1 or x=2x=-3-i\text{ or }x=-1\text{ or }x=2
Question 2
9 markschallenging
Which of the following is equal to (1+3i)(2−i)1+i\frac{\left(1+3i\right)\left(2-i\right)}{1+i}?
Show worked solution

Worked solution

  1. Write down the expression

    (1+3i)(2−i)1+i\frac{\left(1+3i\right)\left(2-i\right)}{1+i}

    Expand the numerator, then rationalise the denominator.

  2. Expand the brackets

    (1+3i)(2−i)=−3i2+5i+2\left(1+3i\right)\left(2-i\right)=-3i^{2}+5i+2

    Multiply each term of the first bracket by each term of the second.

  3. Use i2=−1i^{2}=-1 to remove the i2i^{2} term

    −3i2+5i+2=5+5i-3i^{2}+5i+2=5+5i

    The i2i^{2} term is real, so it changes the real part.

  4. Write down the conjugate of the denominator

    1+i‾=1−i\overline{1+i}=1-i

    The conjugate reverses the sign of the imaginary part.

  5. Multiply numerator and denominator by that conjugate

    5+5i1+i=(5+5i)(1−i)(1+i)(1−i)\frac{5+5i}{1+i}=\frac{\left(5+5i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}

    This does not change the value of the fraction, but it will clear ii from the bottom.

  6. Expand the numerator

    (5+5i)(1−i)=5−5i2\left(5+5i\right)\left(1-i\right)=5-5i^{2}

    Multiply out and keep the i2i^{2} term for the moment.

  7. Simplify the numerator using i2=−1i^{2}=-1

    (5+5i)(1−i)=10\left(5+5i\right)\left(1-i\right)=10

    The numerator is now a single complex number.

  8. Expand the denominator

    (1+i)(1−i)=1−i2\left(1+i\right)\left(1-i\right)=1-i^{2}

    The cross terms in ii cancel because the two brackets are conjugates.

  9. Simplify the denominator

    (1+i)(1−i)=2\left(1+i\right)\left(1-i\right)=2

    A complex number times its conjugate is the real number a2+b2a^{2}+b^{2}.

  10. Divide the real and imaginary parts by the real denominator

    102=5\frac{10}{2}=5

    Splitting the fraction gives the answer in the form a+bia+bi.

  11. Check by multiplying the answer back by the denominator

    (5)(1+i)=5+5i\left(5\right)\left(1+i\right)=5+5i

    Recovering the original numerator confirms the division.

  12. Recall the defining property of ii

    i2=−1i^{2}=-1

    Every simplification of a complex product rests on this single fact.

  13. Recall the definition of a complex number

    z=a+bi,a,b∈Rz=a+bi,\quad a,b\in\mathbb{R}

    aa is the real part and bb is the imaginary part of zz.

  14. Recall the complex conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    The conjugate is obtained by reversing the sign of the imaginary part.

  15. State the answer in the form a+bia+bi

    (1+3i)(2−i)1+i=5\frac{\left(1+3i\right)\left(2-i\right)}{1+i}=5

    This is the value of the expression.

Answer
55
Question 3
9 markschallenging
Which of the following is equal to (1+i)8\left(1+i\right)^{8}?
Show worked solution

Worked solution

  1. Write the power as a repeated product

    (1+i)8=(1+i)⋯(1+i)⏟8 factors\left(1+i\right)^{8}=\underbrace{\left(1+i\right)\cdots\left(1+i\right)}_{8\ \text{factors}}

    Each factor is the same complex number.

  2. Work out (1+i)2\left(1+i\right)^{2}

    (1+i)2=(1+i)(1+i)=i2+2i+1=2i\left(1+i\right)^{2}=\left(1+i\right)\left(1+i\right)=i^{2}+2i+1=2i

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  3. Work out (1+i)3\left(1+i\right)^{3}

    (1+i)3=(2i)(1+i)=2i2+2i=−2+2i\left(1+i\right)^{3}=\left(2i\right)\left(1+i\right)=2i^{2}+2i=-2+2i

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  4. Work out (1+i)4\left(1+i\right)^{4}

    (1+i)4=(−2+2i)(1+i)=2i2−2=−4\left(1+i\right)^{4}=\left(-2+2i\right)\left(1+i\right)=2i^{2}-2=-4

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  5. Work out (1+i)5\left(1+i\right)^{5}

    (1+i)5=(−4)(1+i)=−4i−4=−4−4i\left(1+i\right)^{5}=\left(-4\right)\left(1+i\right)=-4i-4=-4-4i

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  6. Work out (1+i)6\left(1+i\right)^{6}

    (1+i)6=(−4−4i)(1+i)=−4i2−8i−4=−8i\left(1+i\right)^{6}=\left(-4-4i\right)\left(1+i\right)=-4i^{2}-8i-4=-8i

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  7. Work out (1+i)7\left(1+i\right)^{7}

    (1+i)7=(−8i)(1+i)=−8i2−8i=8−8i\left(1+i\right)^{7}=\left(-8i\right)\left(1+i\right)=-8i^{2}-8i=8-8i

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  8. Work out (1+i)8\left(1+i\right)^{8}

    (1+i)8=(8−8i)(1+i)=8−8i2=16\left(1+i\right)^{8}=\left(8-8i\right)\left(1+i\right)=8-8i^{2}=16

    Multiply the previous power by 1+i1+i and use i2=−1i^{2}=-1.

  9. Recall the defining property of ii

    i2=−1i^{2}=-1

    Every simplification of a complex product rests on this single fact.

  10. Recall the definition of a complex number

    z=a+bi,a,b∈Rz=a+bi,\quad a,b\in\mathbb{R}

    aa is the real part and bb is the imaginary part of zz.

  11. Recall the complex conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    The conjugate is obtained by reversing the sign of the imaginary part.

  12. Recall that a number times its conjugate is real

    (a+bi)(a−bi)=a2+b2\left(a+bi\right)\left(a-bi\right)=a^{2}+b^{2}

    This is why multiplying by the conjugate clears ii from a denominator.

  13. Recall the rule for adding complex numbers

    (a+bi)+(c+di)=(a+c)+(b+d)i\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i

    Real parts add to real parts and imaginary parts to imaginary parts.

  14. Recall the rule for subtracting complex numbers

    (a+bi)−(c+di)=(a−c)+(b−d)i\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i

    Subtraction also acts separately on the two components.

  15. Recall the rule for multiplying complex numbers

    (a+bi)(c+di)=(ac−bd)+(ad+bc)i\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i

    This follows from expanding the brackets and using i2=−1i^{2}=-1.

  16. Recall the method for dividing complex numbers

    zw=zw‾ww‾\frac{z}{w}=\frac{z\overline{w}}{w\overline{w}}

    Multiplying top and bottom by the conjugate of the denominator makes it real.

  17. State the answer in the form a+bia+bi

    (1+i)8=16\left(1+i\right)^{8}=16

    This is the value of the expression.

Answer
1616
Question 4
9 markschallenging
Which of the following gives the roots of the equation x4−x3+x2+9x−10=0x^{4}-x^{3}+x^{2}+9x-10=0?
Show worked solution

Worked solution

  1. Write down the equation to be solved

    x4−x3+x2+9x−10=0x^{4}-x^{3}+x^{2}+9x-10=0

    There are 44 roots in total, counted with multiplicity.

  2. Look for a real root by the factor theorem

    p(1)=0p\left(1\right)=0

    Since this is zero, x−(1)x-\left(1\right) is a factor.

  3. Divide out the factor x−(1)x-\left(1\right)

    x4−x3+x2+9x−10=(x−(1))(x3+x+10)x^{4}-x^{3}+x^{2}+9x-10=\left(x-\left(1\right)\right)\left(x^{3}+x+10\right)

    Algebraic division (or comparing coefficients) gives the other factor.

  4. Look for a real root by the factor theorem

    p(−2)=0p\left(-2\right)=0

    Since this is zero, x−(−2)x-\left(-2\right) is a factor.

  5. Divide out the factor x−(−2)x-\left(-2\right)

    x3+x+10=(x−(−2))(x2−2x+5)x^{3}+x+10=\left(x-\left(-2\right)\right)\left(x^{2}-2x+5\right)

    Algebraic division (or comparing coefficients) gives the other factor.

  6. Identify the coefficients

    a=1,b=−2,c=5a=1,\quad b=-2,\quad c=5

    Compare with the general quadratic ax2+bx+c=0ax^{2}+bx+c=0.

  7. Calculate the discriminant

    b2−4ac=(−2)2−4(1)(5)=−16b^{2}-4ac=\left(-2\right)^{2}-4\left(1\right)\left(5\right)=-16

    The discriminant is negative, so the roots are a complex conjugate pair.

  8. Apply the quadratic formula

    x=−(−2)±−162x=\frac{-\left(-2\right)\pm\sqrt{-16}}{2}

    Substitute the coefficients into x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.

  9. Write the square root of the negative number in terms of ii

    −16=4i\sqrt{-16}=4i

    Use −k=ik\sqrt{-k}=i\sqrt{k}.

  10. Simplify the surd expression

    x=2±4i2x=\frac{2\pm4i}{2}

    Both parts of the numerator are divided by 22.

  11. Write down the two roots

    x=1+2i or x=1−2ix=1+2i\text{ or }x=1-2i

    The roots form a conjugate pair, as they must for real coefficients.

  12. Check the sum of the roots

    α+β=2=−ba=2\alpha+\beta=2=-\frac{b}{a}=2

    The sum of the roots of x2+bx+c=0x^{2}+bx+c=0 is −b-b.

  13. Check the product of the roots

    αβ=5=ca=5\alpha\beta=5=\frac{c}{a}=5

    The product of the roots of x2+bx+c=0x^{2}+bx+c=0 is cc.

  14. Substitute the first root back into the quadratic

    −3+4i+(−2−4i)+(5)=0-3+4i+\left(-2-4i\right)+\left(5\right)=0

    The three terms cancel exactly, confirming the root.

  15. Substitute the second root back into the quadratic

    −3−4i+(−2+4i)+(5)=0-3-4i+\left(-2+4i\right)+\left(5\right)=0

    The conjugate root also satisfies the equation.

  16. State the complete solution set

    x4−x3+x2+9x−10=0⇒x=−2 or x=1−2i or x=1 or x=1+2ix^{4}-x^{3}+x^{2}+9x-10=0\Rightarrow x=-2\text{ or }x=1-2i\text{ or }x=1\text{ or }x=1+2i

    All 44 roots of the equation are now known.

Answer
x=−2 or x=1−2i or x=1 or x=1+2ix=-2\text{ or }x=1-2i\text{ or }x=1\text{ or }x=1+2i
Question 5
9 markschallenging
Given that x=1+2ix=1+2i is a root of the equation x4−6x3+8x2−10x−25=0x^{4}-6x^{3}+8x^{2}-10x-25=0, find the remaining roots.
Show worked solution

Worked solution

  1. Write down what is given

    p(x)=x4−6x3+8x2−10x−25,x=1+2i is a rootp\left(x\right)=x^{4}-6x^{3}+8x^{2}-10x-25,\qquad x=1+2i\ \text{is a root}

    The coefficients of pp are real, which is the key to the whole question.

  2. Apply the conjugate root theorem

    x=1+2i is a root⇒x=1−2i is a rootx=1+2i\ \text{is a root}\Rightarrow x=1-2i\ \text{is a root}

    Non-real roots of a real polynomial always come in conjugate pairs.

  3. Form the quadratic factor from the conjugate pair

    (x−(1+2i))(x−(1−2i))=x2−2x+5\left(x-\left(1+2i\right)\right)\left(x-\left(1-2i\right)\right)=x^{2}-2x+5

    The pair multiplies out to a quadratic with real coefficients.

  4. Check the sum and product of that pair

    α+β=2,αβ=5\alpha+\beta=2,\quad\alpha\beta=5

    These reproduce the coefficients of the quadratic factor.

  5. Divide p(x)p\left(x\right) by the quadratic factor

    x4−6x3+8x2−10x−25=(x2−2x+5)(x2−4x−5)x^{4}-6x^{3}+8x^{2}-10x-25=\left(x^{2}-2x+5\right)\left(x^{2}-4x-5\right)

    Algebraic division leaves the remaining factor.

  6. Verify the factorisation by expanding

    (x2−2x+5)(x2−4x−5)=x4−6x3+8x2−10x−25\left(x^{2}-2x+5\right)\left(x^{2}-4x-5\right)=x^{4}-6x^{3}+8x^{2}-10x-25

    Expanding the product returns the original polynomial.

  7. Factorise the remaining quadratic

    x2−4x−5=x2−4x−5x^{2}-4x-5=x^{2}-4x-5

    The remaining quadratic has real roots.

  8. Solve the remaining quadratic

    x=−1 or x=5x=-1\text{ or }x=5

    These are the two real roots.

  9. Check the root x=−1x=-1 in p(x)p\left(x\right)

    p(−1)=0p\left(-1\right)=0

    Substituting the root into the polynomial gives zero.

  10. Check the root x=1−2ix=1-2i in p(x)p\left(x\right)

    p(1−2i)=0p\left(1-2i\right)=0

    Substituting the root into the polynomial gives zero.

  11. Check the root x=5x=5 in p(x)p\left(x\right)

    p(5)=0p\left(5\right)=0

    Substituting the root into the polynomial gives zero.

  12. Recall the defining property of ii

    i2=−1i^{2}=-1

    Every simplification of a complex product rests on this single fact.

  13. Recall the definition of a complex number

    z=a+bi,a,b∈Rz=a+bi,\quad a,b\in\mathbb{R}

    aa is the real part and bb is the imaginary part of zz.

  14. Recall the complex conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    The conjugate is obtained by reversing the sign of the imaginary part.

  15. State the remaining roots

    x≠1+2i:x=−1 or x=1−2i or x=5x\neq1+2i:\quad x=-1\text{ or }x=1-2i\text{ or }x=5

    These are the other 33 roots of the equation.

Answer
x=−1 or x=1−2i or x=5x=-1\text{ or }x=1-2i\text{ or }x=5

Unlock 29 more Complex numbers: arithmetic questions

Create a free account to work through every Further Maths Complex numbers: arithmetic question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Complex numbers: arithmetic practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsFurther calculusHyperbolic functions