Video: Solving Quadratic Equations with Complex Coefficients

Solve (1 + 2𝑖)𝑧² βˆ’ 3 + 𝑖 = 0. Round your answers to three significant figures.


Video Transcript

Solve one plus two 𝑖 times 𝑧 squared minus three plus 𝑖 equals zero. Round your answers to three significant figures.

As there’s no term involving 𝑧, it seems a bit silly to break out the quadratic formula. What we can do instead is to subtract negative three plus 𝑖 from both sides and then divide both sides by one plus two 𝑖 to find 𝑧 squared equals three minus 𝑖 over one plus two 𝑖.

Now we just need to find the square root of this complex number. But first, we multiply numerator and denominator by the complex conjugate of the denominator. We can distribute and then simplify. And we get a fifth minus seven-fifths 𝑖.

Now how do we find the square root of this number? Well, we can write it in polar form and then apply de Moivre’s theorem for roots. What’s the modulus of this complex number? We find that it is root two. As our complex number is in the fourth quadrant, its argument is the arctan of its imaginary part over its real part. We find that its argument is arctan of negative seven. And so we can write our complex number in polar form.

De Moivre’s theorem for roots gives us the 𝑛th roots of a complex number in polar form. We’re looking for square roots, and so 𝑛 is two. Applying this formula to our example, where π‘Ÿ is root two and πœƒ is arctan negative seven, we get the following. Putting these into a calculator, we get 𝑧 equals 0.898 minus 0.779𝑖 and 𝑧 equals negative 0.898 plus 0.77𝑖 to three significant figures.

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