Explainer: Quadratic Equations with Complex Coefficients

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ƒ(οŠ°οŽ„)ο‘΅οŽ‘. Considering the second root 𝛽, we have 𝛽=βˆšπ‘Ÿπ‘’=βˆšπ‘Ÿπ‘’.(οŠ°οŽ„)ο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.