McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 3

Lesson 3 · Quadratic equations and square roots

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 4 · Week 7
Learning intentionTo solve quadratic equations whose solutions are complex, and to find the square roots of a complex number.
Success criteria · I can…
  • solve quadratics with real coefficients and Δ<0\Delta<0
  • explain why their complex roots are conjugate pairs, and form a quadratic from one root
  • find the two square roots of a+iba+ib by writing (x+iy)2=a+ib(x+iy)^2=a+ib
  • solve quadratics with complex coefficients

Lesson flow

LaunchWarm-up · Retrieval from Lessons 1 and 2Slide 2ExamplesExample 1 · Quadratic formulaSlide 3ExamplesExample 2 · Completing the squareSlide 4DefinitionsConjugate pairsSlide 5HSC-styleHSC-style · Coefficients from a rootSlide 6DefinitionsSquare roots of a complex numberSlide 7ExamplesExample 5 · Square rootsSlide 8Key ideaExample 6 · A faster waySlide 9HSC-styleHSC-style · Complex coefficientsSlide 10ExamplesExample 7 · A harder oneSlide 11HSC-styleHSC-style · Using square rootsSlide 12Exit ticketExit ticketSlide 13

Warm-up · Retrieval from Lessons 1 and 2

Slide 2
(a) Simplify i103i^{103}.
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Answer
−i-i
Why
103=4×25+3103=4\times25+3, so i103=i3=−ii^{103}=i^3=-i.
(b) Expand (3−2i)2(3-2i)^2.
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Answer
5−12i5-12i
Why
9−12i+4i2=9−12i−4=5−12i9-12i+4i^2=9-12i-4=5-12i. This expansion step is used for square roots later in the lesson.
(c) Express 2+i1−i\dfrac{2+i}{1-i} in the form a+iba+ib.
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Answer
1+3i2\dfrac{1+3i}{2}
Why
Multiply by the conjugate of the denominator: (2+i)(1+i)(1−i)(1+i)=2+2i+i−12=1+3i2\dfrac{(2+i)(1+i)}{(1-i)(1+i)}=\dfrac{2+2i+i-1}{2}=\dfrac{1+3i}{2}.
(d) Solve x2+16=0x^2+16=0.
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Answer
x=±4ix=\pm4i
Why
x2=−16=16i2x^2=-16=16i^2, so x=±4ix=\pm4i.

Example 1 · Quadratic formula

Slide 3

Consider x2+2x+4=0x^2+2x+4=0.

(a) Find the sum and the product of the roots.
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Answer
Sum =−2=-2, product =4=4.
Why
For x2+2x+4=0x^2+2x+4=0, the sum of the roots is −ba=−2-\frac{b}{a}=-2 and the product is ca=4\frac{c}{a}=4.
(b) Use the quadratic formula to solve for xx.
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Answer
x=−1±i3x=-1\pm i\sqrt3
Why
x=−2±4−162=−2±−122=−2±23 i2=−1±3 ix=\frac{-2\pm\sqrt{4-16}}{2}=\frac{-2\pm\sqrt{-12}}{2}=\frac{-2\pm2\sqrt3\,i}{2}=-1\pm\sqrt3\,i. Check: the sum is −2-2 and the product is 1+3=41+3=4. Notice that the two roots are conjugates of each other.

Example 2 · Completing the square

Slide 4
Solve z2+8z+25=0z^2+8z+25=0 by completing the square.
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Answer
z=−4±3iz=-4\pm3i
Why
(z+4)2+9=0(z+4)^2+9=0, so (z+4)2=−9(z+4)^2=-9, z+4=±3iz+4=\pm3i and z=−4±3iz=-4\pm3i. Note that a sum of squares factorises over C\mathbb{C}: z2+λ2=(z+iλ)(z−iλ)z^2+\lambda^2=(z+i\lambda)(z-i\lambda).

Conjugate pairs

Slide 5
Theorem
Real coefficients give conjugate roots

If az2+bz+c=0az^2+bz+c=0 with a,b,c∈Ra,b,c\in\mathbb{R} has a non-real root α\alpha, then the other root is α‾\conj{\alpha}.

Since α+α‾=2Re⁡(α)\alpha+\conj{\alpha}=2\Re(\alpha) and αα‾=∣α∣2\alpha\conj{\alpha}=|\alpha|^2, a monic quadratic with root α\alpha is

z2−2Re⁡(α) z+∣α∣2=0z^2-2\Re(\alpha)\,z+|\alpha|^2=0

Why it works: the roots are −b±i4ac−b22a\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}, which are conjugates when a,b,ca,b,c are real.

Example 3

Find a quadratic equation with real coefficients that has w=2−iw=2-i as one root.
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Answer
z2−4z+5=0z^2-4z+5=0
Why
The other root is w‾=2+i\conj{w}=2+i. Sum =4=4, product =4+1=5=4+1=5. Equivalently (z−(2−i))(z−(2+i))=(z−2)2+1=0(z-(2-i))(z-(2+i))=(z-2)^2+1=0.

HSC-style · Coefficients from a root

Slide 6
Adapted from 2020 HSC Q2The equation z2+pz+q=0z^2+pz+q=0, where pp and qq are real, has 3+i3+i as a root. Find pp and qq.
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Answer
p=−6p=-6 and q=10q=10.
Why
The other root is 3−i3-i. Sum of roots =6=−p=6=-p, so p=−6p=-6. Product =(3+i)(3−i)=10=q=(3+i)(3-i)=10=q. A common slip is forgetting the minus sign in −p-p and giving p=6p=6.

Square roots of a complex number

Slide 7
Key idea
Finding the square roots of a+iba+ib

Let (x+iy)2=a+ib(x+iy)^2=a+ib, where x,y∈Rx,y\in\mathbb{R}. Expand and equate real and imaginary parts:

x2−y2=aand2xy=bx^2-y^2=a\qquad\text{and}\qquad 2xy=b

Every non-zero complex number has two square roots, one the negative of the other.

Watch out

Write (x+iy)2=a+ib(x+iy)^2=a+ib, not x+iy=a+ibx+iy=\sqrt{a+ib}. Say “find the square roots of a+iba+ib”, in the plural.

Example 4

Find the square roots of −9-9.
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Answer
±3i\pm3i
Why
(±3i)2=9i2=−9(\pm3i)^2=9i^2=-9. With the method: x2−y2=−9x^2-y^2=-9 and 2xy=02xy=0 give x=0x=0, y=±3y=\pm3.

Example 5 · Square roots

Slide 8
Find the square roots of 7+24i7+24i.
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Answer
±(4+3i)\pm(4+3i)
Why
Let (x+iy)2=7+24i(x+iy)^2=7+24i. Then x2−y2=7x^2-y^2=7 (1) and 2xy=242xy=24, so y=12xy=\frac{12}{x} (2). Substitute (2) into (1): x2−144x2=7x^2-\frac{144}{x^2}=7, so x4−7x2−144=0x^4-7x^2-144=0, that is (x2−16)(x2+9)=0(x^2-16)(x^2+9)=0. Since xx is real, x2=16x^2=16, so x=±4x=\pm4 and y=±3y=\pm3 (same sign). Check: (4+3i)2=16−9+24i(4+3i)^2=16-9+24i.

Example 6 · A faster way

Slide 9
Key idea
Use the modulus as a third equation

Taking moduli of (x+iy)2=a+ib(x+iy)^2=a+ib gives x2+y2=a2+b2x^2+y^2=\sqrt{a^2+b^2}. Add and subtract this with x2−y2=ax^2-y^2=a to find x2x^2 and y2y^2.

Signs: from 2xy=b2xy=b, xx and yy have the same sign if b>0b>0 and opposite signs if b<0b<0.

Example 6

Find the square roots of each number.

(a) −5+12i-5+12i
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Answer
±(2+3i)\pm(2+3i)
Why
x2−y2=−5x^2-y^2=-5 and x2+y2=13x^2+y^2=13, so x2=4x^2=4, y2=9y^2=9. Since b=12>0b=12>0 the signs match: ±(2+3i)\pm(2+3i).
2021 HSC Q11(d)(i)(b) −i-i
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Answer
±12(1−i)\pm\dfrac{1}{\sqrt2}(1-i)
Why
x2−y2=0x^2-y^2=0 and x2+y2=1x^2+y^2=1, so x2=y2=12x^2=y^2=\frac12. Since b=−1<0b=-1<0 the signs are opposite: ±12(1−i)\pm\frac{1}{\sqrt2}(1-i). Check: (1−i)2=−2i(1-i)^2=-2i, and 12(−2i)=−i\frac12(-2i)=-i.

HSC-style · Complex coefficients

Slide 10

For a,b,c∈Ca,b,c\in\mathbb{C} the quadratic formula z=−b±Δ2az=\dfrac{-b\pm\sqrt{\Delta}}{2a} still works, where ±Δ\pm\sqrt\Delta means the two square roots of Δ=b2−4ac\Delta=b^2-4ac. The roots are usually not conjugates now.

2020 HSC Q11(e)Solve z2+3z+(3−i)=0z^2+3z+(3-i)=0.
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Answer
z=−1+iz=-1+i or z=−2−iz=-2-i.
Why
Δ=9−4(3−i)=−3+4i\Delta=9-4(3-i)=-3+4i. For the square roots: x2−y2=−3x^2-y^2=-3 and x2+y2=5x^2+y^2=5 give x2=1x^2=1, y2=4y^2=4, same sign, so ±(1+2i)\pm(1+2i). Then z=−3±(1+2i)2z=\frac{-3\pm(1+2i)}{2}. Check: the sum is −3-3 and the product is (−1+i)(−2−i)=2+i−2i+1=3−i(-1+i)(-2-i)=2+i-2i+1=3-i.

Example 7 · A harder one

Slide 11
Solve z2+2(1−i3)z+2+2i3=0z^2+2\left(1-i\sqrt3\right)z+2+2i\sqrt3=0.
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Answer
z=(−1+2)+(3−6) iz=(-1+\sqrt2)+(\sqrt3-\sqrt6)\,i or z=(−1−2)+(3+6) iz=(-1-\sqrt2)+(\sqrt3+\sqrt6)\,i.
Why
Δ=4(1−i3)2−4(2+2i3)=4(−2−23 i)−8−83 i=−16−163 i=16(−1−3 i)\Delta=4(1-i\sqrt3)^2-4(2+2i\sqrt3)=4(-2-2\sqrt3\,i)-8-8\sqrt3\,i=-16-16\sqrt3\,i=16(-1-\sqrt3\,i). Square roots of −1−3 i-1-\sqrt3\,i: x2−y2=−1x^2-y^2=-1 and x2+y2=2x^2+y^2=2 give x2=12x^2=\frac12, y2=32y^2=\frac32; b<0b<0 so the signs are opposite: ±(12−32 i)\pm\left(\frac{1}{\sqrt2}-\sqrt{\tfrac32}\,i\right). So Δ=±(22−26 i)\sqrt\Delta=\pm(2\sqrt2-2\sqrt6\,i) and z=−1+3 i±(2−6 i)z=-1+\sqrt3\,i\pm(\sqrt2-\sqrt6\,i). Never leave a complex number under a square root sign: find its square roots first.

HSC-style · Using square roots

Slide 12
2021 HSC Q11(d)(i) Find the square roots of −i-i. (ii) Hence solve z2+2z+(1+i)=0z^2+2z+(1+i)=0.
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Answer
(i) ±12(1−i)\pm\dfrac{1}{\sqrt2}(1-i). (ii) z=−1+12−12 iz=-1+\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt2}\,i or z=−1−12+12 iz=-1-\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}\,i.
Why
(i) is Example 6(b). (ii) Δ=4−4(1+i)=−4i\Delta=4-4(1+i)=-4i, so Δ=±2⋅12(1−i)=±2(1−i)\sqrt\Delta=\pm2\cdot\frac{1}{\sqrt2}(1-i)=\pm\sqrt2(1-i) and z=−2±2(1−i)2=−1±12(1−i)z=\frac{-2\pm\sqrt2(1-i)}{2}=-1\pm\frac{1}{\sqrt2}(1-i). Completing the square gives the same: (z+1)2=−i(z+1)^2=-i.

Exit ticket

Slide 13
1. Find the square roots of −3−4i-3-4i.
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Answer
±(1−2i)\pm(1-2i)
Why
x2−y2=−3x^2-y^2=-3 and x2+y2=5x^2+y^2=5, so x2=1x^2=1, y2=4y^2=4. Since b=−4<0b=-4<0 the signs are opposite. Check: (1−2i)2=1−4i−4=−3−4i(1-2i)^2=1-4i-4=-3-4i.
2. Hence solve z2−3z+(3+i)=0z^2-3z+(3+i)=0.
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Answer
z=2−iz=2-i or z=1+iz=1+i.
Why
Δ=9−4(3+i)=−3−4i\Delta=9-4(3+i)=-3-4i, so Δ=±(1−2i)\sqrt\Delta=\pm(1-2i) and z=3±(1−2i)2z=\frac{3\pm(1-2i)}{2}. Check: the sum is 33 and the product is (2−i)(1+i)=3+i(2-i)(1+i)=3+i.