Lesson Explainer: Quadratic Equations with Complex Coefficients Mathematics

In this explainer, we will learn how to solve quadratic equations with complex coefficients using the quadratic formula.

From the fundamental theorem of algebra, we know that any quadratic will have two roots. For quadratic equations with real coefficients, the conjugate root theorem tells us that if it has any nonreal root, its roots will be a complex conjugate pair, whereas if it has real roots, it could either have two distinct real roots or a single repeated root. To distinguish between these three different cases, we use the discriminant.

Discriminant

The discriminant of a quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0 is defined as π‘βˆ’4π‘Žπ‘οŠ¨. Often Ξ” is used to denote the discriminant.

Using the discriminant, we identify the three different cases of quadratic equations as follows.

  1. Positive discriminant: π‘βˆ’4π‘Žπ‘>0, two real roots
  2. Zero discriminant: π‘βˆ’4π‘Žπ‘=0, one repeated real root
  3. Negative discriminant: π‘βˆ’4π‘Žπ‘<0, complex conjugate roots

The graphs below depict each case.

In this explainer, we will relax the condition that the quadratic equation has real coefficients and explore what we can conclude about the roots of quadratic equations with complex coefficients.

Let us consider how to solve a general quadratic equation 0=π‘Žπ‘§+𝑏𝑧+𝑐, where π‘Žβ‰ 0. We start by dividing the equation by π‘Ž, which gives 0=𝑧+π‘π‘Žπ‘§+π‘π‘Ž.

We can now complete the square 0=𝑧+𝑏2π‘Žο‰βˆ’π‘4π‘Ž+π‘π‘Ž=𝑧+𝑏2π‘Žο‰βˆ’π‘βˆ’4π‘Žπ‘4π‘Ž.

Rearranging, we get

𝑧+𝑏2π‘Žο‰=π‘βˆ’4π‘Žπ‘4π‘Ž.(1)

All of the steps we have taken up to this point are equally valid whether π‘Ž, 𝑏, and 𝑐 are real or imaginary. The next step, where we take the square root, is something we might need to be careful with, since taking roots of a complex number returns multiple values. However, let us consider the case of a general complex number 𝑀=π‘Ÿπ‘’οƒοΌ. Using de Moivre’s theorem, the two possible square roots of this number are 𝛼=βˆšπ‘Ÿπ‘’οƒο‘΅οŽ‘ and 𝛽=βˆšπ‘Ÿπ‘’οƒο€ΌοŠ°οŽ„οˆο‘΅οŽ‘. Considering the second root 𝛽, we have 𝛽=βˆšπ‘Ÿπ‘’=βˆšπ‘Ÿπ‘’.οƒο€ΌοŠ°οŽ„οˆοƒοŠ°οƒοŽ„ο‘΅οŽ‘ο‘΅οŽ‘

Using the properties of exponential functions, we can express this as 𝛽=βˆšπ‘Ÿπ‘’π‘’.οƒοƒοŽ„ο‘΅οŽ‘

However, courtesy of the Euler’s identity, we know 𝑒=βˆ’1οƒοŽ„; therefore, 𝛽=βˆ’βˆšπ‘Ÿπ‘’=βˆ’π›Ό.οƒο‘΅οŽ‘

Therefore, in the same way that we take both the positive and negative roots for real numbers, we can find a root of a complex number using de Moirve’s theorem, and then the second root will be its negative. Hence, returning to equation (1), we can take square roots and consider both the positive and negative values, which gives 𝑧+𝑏2π‘Ž=Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Therefore, 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Hence, even for quadratics with complex coefficients, we can use the quadratic formula, although we might need to appeal to de Moirve’s theorem to calculate the value of the square root of the discriminant for nonreal numbers.

In the first couple of examples, we will consider the special case where, in spite of the fact that we have some nonreal coefficients, the discriminant is still a real number.

Example 1: Nonreal Quadratic Equations with Real Negative Discriminants

Solve 3𝑧+5π‘–π‘§βˆ’2=0.

Answer

Using the quadratic formula 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, we have 𝑧=βˆ’5𝑖±(5𝑖)βˆ’4Γ—3(βˆ’2)2Γ—3=βˆ’5π‘–Β±βˆšβˆ’25+246=βˆ’5𝑖±𝑖6.

Hence, 𝑧=βˆ’23𝑖 and 𝑧=βˆ’π‘– are the solutions of the equation.

Notice that in the previous example, in spite of the fact that we had a negative discriminant, the two solutions are not a complex conjugate pair. Hence, the rule regarding a negative discriminant does not simply generalize to equations with complex coefficients.

Example 2: Nonreal Quadratic Equations with Real Positive Discriminants

Solve 𝑧+(2+𝑖)𝑧+𝑖=0.

Answer

Using the quadratic formula 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, we have 𝑧=βˆ’(2+𝑖)±(2+𝑖)βˆ’4𝑖2=βˆ’2βˆ’π‘–Β±βˆš3+4π‘–βˆ’4𝑖2=βˆ’2βˆ’π‘–Β±βˆš32.

Hence, 𝑧=βˆ’2+√32βˆ’π‘–2 and 𝑧=βˆ’2βˆ’βˆš32βˆ’π‘–2 are the solutions of the equation.

In the previous example, we had a positive discriminant; however, the two solutions are not real. Therefore, the rule regarding the positive discriminant does not simply generalize to equations with complex coefficients.

Example 3: Nonreal Quadratic Equations with Real Discriminants

Solve (2+3𝑖)𝑧+4π‘§βˆ’6𝑖+4=0.

Answer

Applying the quadratic formula 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, we have 𝑧=βˆ’4±√4βˆ’4(2+3𝑖)(4βˆ’6𝑖)2(2+3𝑖)=βˆ’4±√4(4βˆ’(2+3𝑖)(4βˆ’6𝑖))2(2+3𝑖).

Taking the 4 out of the square root, we have 𝑧=βˆ’4Β±2√4βˆ’(2+3𝑖)(4βˆ’6𝑖)2(2+3𝑖).

Canceling the factor of 2 in the numerator and denominator, we have 𝑧=βˆ’2±√4βˆ’(2+3𝑖)(4βˆ’6𝑖)2+3𝑖.

Notice also that 4βˆ’6𝑖=2(2βˆ’3𝑖). Hence, (2+3𝑖)(4βˆ’6𝑖) is twice the complex number 2+3𝑖 times its conjugate. Therefore, (2+3𝑖)(4βˆ’6𝑖)=2ο€Ή2+3=26, and thus 𝑧=βˆ’2±√4βˆ’262+3𝑖=βˆ’2Β±π‘–βˆš222+3𝑖.

Multiplying both the numerator and denominator by the complex conjugate of the denominator gives 𝑧=ο€»βˆ’2Β±π‘–βˆš22(2βˆ’3𝑖)(2+3𝑖)(2βˆ’3𝑖)=βˆ’4+6𝑖±3√22+2π‘–βˆš2213.

Hence, 𝑧=βˆ’4+3√2213+6+2√2213𝑖 and 𝑧=βˆ’4βˆ’3√2213+6βˆ’2√2213𝑖 are the solutions of the equation.

Notice that, unlike the first two examples, the two roots in the previous examples do not lie on a vertical or horizontal line in the complex plane. The following figures represent the graph of the modulus of each equation plotted above the complex plane. The two points where the graph meets the complex plane are the roots of the equations.

Example 4: Nonreal Quadratic Equations with Zero Discriminants

Solve π‘§βˆ’(4+4𝑖)𝑧+8𝑖=0.

Answer

Using the quadratic formula 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, we have 𝑧=4+4𝑖±(βˆ’(4+4𝑖))βˆ’4(8𝑖)2=4+4π‘–Β±βˆš32π‘–βˆ’32𝑖2=2+2𝑖.

Hence, the equation has a single repeated root of 𝑧=2+2𝑖.

The previous example demonstrates that it is possible to have repeated nonreal roots. Furthermore, we see that when this happened, the discriminant was zero. By simply considering the quadratic formula, we can see that the rule regarding a determinant of zero generalizes to the case where we have nonreal coefficients. This is in fact one of the key properties of the discriminant: that it is zero if, and only if, the equation has a repeated root.

In the following examples, we will consider the case where the discriminant is a general complex number. We will begin with a simple example where the coefficient of 𝑧 is zero, which of course is equivalent to simply finding the square roots of a complex number.

Example 5: Solving Quadratic Equations by Taking the Square Root of Complex Numbers

Solve (1+2𝑖)π‘§βˆ’3+𝑖=0. Round your answers to three significant figures.

Answer

Adding 3βˆ’π‘– to both sides of the equation gives (1+2𝑖)𝑧=3βˆ’π‘–.

We now divide by (1+2𝑖) to isolate π‘§οŠ¨ on the left-hand side of the equation: 𝑧=3βˆ’π‘–1+2𝑖.

By multiplying the numerator and denominator by the complex conjugate of the denominator, we can simplify the complex fraction as follows: 𝑧=(3βˆ’π‘–)(1βˆ’2𝑖)(1+2𝑖)(1βˆ’2𝑖)=3βˆ’6π‘–βˆ’π‘–βˆ’25=15βˆ’75𝑖.

To find the square roots of this complex number, we need to calculate its modulus and argument. The modulus is given by ο„Ÿο€Ό15+ο€Όβˆ’75=√2. Since it lies in the fourth quadrant, its argument is given by πœƒ=ο‚βˆ’οŽ=(βˆ’7).arctanarctan

Applying de Moivre’s theorem, the two square roots are given by ±√2ο€Ό12πœƒ+𝑖12πœƒοˆ.cossin

Substituting in the value of πœƒ and rounding our answers to three significant figures, we have 𝑧=0.898βˆ’0.779𝑖(3s.f.) and 𝑧=βˆ’0.898+0.779𝑖(3s.f.).

Example 6: Quadratic Equations with Complex Discriminants

Solve 𝑧+(2βˆ’2𝑖)π‘§βˆ’(7+26𝑖)=0.

Answer

Using the quadratic formula 𝑧=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž, we have

𝑧=βˆ’(2βˆ’2𝑖)±(2βˆ’2𝑖)+4(7+26𝑖)2=βˆ’(2βˆ’2𝑖)Β±βˆšβˆ’8𝑖+28+104𝑖2=βˆ’(2βˆ’2𝑖)±√28+96𝑖2.(2)

To evaluate the square root of 28+96𝑖, we first need to calculate the modulus and argument. The modulus is given by √28+96=100. Since the complex number lies in the first quadrant, its argument is given by πœƒ=9628=247arctanarctan. Applying de Moivre’s theorem, the two square roots are given by ±√100ο€Ό12πœƒ+𝑖12πœƒοˆ=Β±10ο€Ό12πœƒ+𝑖12πœƒοˆ.cossincossin

Substituting in the value of πœƒ, we find cos12πœƒ=0.8 and sin12πœƒ=0.6. Hence, the two square roots are given by 8+6𝑖 and βˆ’8βˆ’6𝑖. Substituting these value back into (2), we have 𝑧=βˆ’(2βˆ’2𝑖)Β±(8+6𝑖)2=βˆ’1+𝑖±(4+3𝑖).

Hence, the two roots of the equation are 𝑧=3+4𝑖 and 𝑧=βˆ’5βˆ’2𝑖.

We finish by considering one final example.

Example 7: Polynomials with Nonreal Coefficients

Solve 𝑧+ο€»6+6π‘–βˆš(3)𝑧+32βˆ’32π‘–βˆš(3)οŠͺ.

Answer

We first notice that we have a quadratic in π‘§οŠ¨. Therefore, we start by making the substitution 𝑀=π‘§οŠ¨, which gives 𝑀+ο€»6+6π‘–βˆš(3)𝑀+32βˆ’32π‘–βˆš(3).

We can solve this using the quadratic formula 𝑀=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘ŽοŠ¨ as follows: 𝑀=βˆ’ο€»6+6π‘–βˆš3ο‡Β±ο„žο€»6+6π‘–βˆš3ο‡βˆ’4ο€»32βˆ’32π‘–βˆš32=βˆ’6βˆ’6π‘–βˆš3Β±ο„βˆ’72+72π‘–βˆš3βˆ’128+128π‘–βˆš32=βˆ’6βˆ’6π‘–βˆš3Β±ο„βˆ’200+200π‘–βˆš32.

Canceling the 2 from the numerator and denominator, we have

𝑀=βˆ’3βˆ’3π‘–βˆš3Β±ο„βˆ’50+50π‘–βˆš3.(3)

To evaluate the square root of βˆ’50+50π‘–βˆš3, we need to first calculate its modulus and argument. The modulus is given by ο„ž(βˆ’50)+ο€»50√3=100. Since this complex number lies in the second quadrant, its argument πœƒ is given by πœƒ=ο€Ώβˆ’50√350+πœ‹=ο€»βˆ’βˆš3+πœ‹=βˆ’πœ‹3+πœ‹=2πœ‹3.arctanarctan

Using de Moivre’s theorem, its square roots are given by ±√100ο€Ό12πœƒ+𝑖12πœƒοˆ=Β±10ο€»ο€»πœ‹3+π‘–ο€»πœ‹3.cossincossin

Converting this back into algebraic form, we have that the two roots are 5+5π‘–βˆš3 and βˆ’5βˆ’5π‘–βˆš3. Substituting these into (3), we get 𝑀=βˆ’3βˆ’3π‘–βˆš3Β±5+5π‘–βˆš3.

Hence, the two roots are 2+2π‘–βˆš3 and βˆ’8βˆ’8π‘–βˆš3. Finally, to find the possible values of 𝑧, we need to take the square roots of each of these complex numbers, starting with 2+2π‘–βˆš3. We can express this in exponential form as 4π‘’ο‘½οŽ’οƒ. Applying de Moivre’s theorem, the two square roots are given by Β±2𝑒οŽ₯ which we can express in algebraic form as √3+𝑖 and βˆ’βˆš3βˆ’π‘–. Similarly for βˆ’8βˆ’8π‘–βˆš3, we can express this in exponential form as 16π‘’οŠ±οƒοŽ‘ο‘½οŽ’. Applying de Moivre’s theorem, the two square roots are given by Β±4π‘’οŠ±οƒο‘½οŽ’, which we can express in algebraic form as 2βˆ’2π‘–βˆš3 and βˆ’2+2π‘–βˆš3.

Therefore, the four roots of the equation are given by 𝑧=√3+𝑖, 𝑧=βˆ’βˆš3βˆ’π‘–, 𝑧=2βˆ’2π‘–βˆš3, and 𝑧=βˆ’2+2π‘–βˆš3.

Key Points

  • We can solve quadratic equations with complex coefficients using the quadratic formula.
  • If the discriminant is zero, the equation has one repeated root. If all the coefficients are real, the root will be real.
  • A general quadratic with complex coefficients can have any combination of real and nonreal roots.
  • If a quadratic has real coefficients, the discriminant is sufficient to completely classify the types of solution.

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