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Complex conjugates and polynomial rootsAQA A-Level Further Maths: Flashcards

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Question

What is the complex conjugate of $3-2i$?

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What is the complex conjugate of 3−2i3-2i?
3+2i3+2i
What is zz∗zz^* for z=x+iyz=x+iy?
x2+y2x^2+y^2, which is real.
State the conjugate root theorem.
If a polynomial has real coefficients, its non-real roots occur in conjugate pairs.
Why does the theorem need real coefficients?
Conjugation changes p(z0)=0p(z_0)=0 into p(z0∗)=0p(z_0^*)=0 only if the coefficients are unchanged by conjugation, i.e. real.
Quadratic with roots a±bia\pm bi?
z2−2az+a2+b2z^2-2az+a^2+b^2
Real quadratic factor with roots 2±i2\pm i?
z2−4z+5z^2-4z+5
How many real roots must a real cubic have?
At least one: 11 or 33.
How many real roots can a real quartic have?
00, 22 or 44.
Steps to solve a cubic given a real root?
Divide by the linear factor (or compare coefficients), then solve the quadratic.
Steps to solve a quartic given a complex root a+bia+bi?
State a−bia-bi, form z2−2az+a2+b2z^2-2az+a^2+b^2, find the other quadratic factor, solve both.
Roots of z2+4=0z^2+4=0?
±2i\pm2i
Solve z2−2z+5=0z^2-2z+5=0.
z=1±2iz=1\pm2i

Exam questions on Complex conjugates and polynomial roots

  1. The cubic equation z3−5z2+17z−13=0z^3-5z^2+17z-13=0 has a real root z=1z=1.
    Explain why the cubic has exactly one real root.2 marks
  2. The cubic equation z3−4z2+6z−4=0z^3-4z^2+6z-4=0 has real coefficients and a root z=1+iz=1+i.
    Show by substitution that 1−i1-i is a root of the equation.2 marks
  3. The quartic f(z)=z4−2z3+9z2−8z+20f(z)=z^4-2z^3+9z^2-8z+20 has z2−2z+5z^2-2z+5 as a factor.
    Find the other quadratic factor of f(z)f(z).3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).