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

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

03:52

Video Transcript

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

We have a couple of ways that we could answer this question. This’s quite clearly a quadratic equation with nonreal coefficients. And as such, we could apply the quadratic formula. Alternatively though, if we notice there is no coefficient of 𝑧, or actually the coefficient of 𝑧 is zero, we see that we can actually simply add three and subtract 𝑖 from both sides to start solving this equation.

And so, our first line of working is one plus two 𝑖 times 𝑧 squared equals three minus 𝑖. Next, we’re going to divide through by the coefficient of 𝑧 squared, by one plus two 𝑖 such that 𝑧 squared is equal to three minus 𝑖 over one plus two 𝑖. Now, this isn’t a particularly nice equation to work with. On the left, we have 𝑧 squared. But also, on the right, we have the quotient of two complex numbers.

So, instead, we recall if we multiply a complex number by its conjugate β€” that is, we change the sign of the imaginary part β€” we end up with a real number. And so, we’re going to multiply both the numerator and denominator of our fraction by the conjugate of the denominator, by one minus two 𝑖. Let’s perform this calculation over here.

We multiply the first term in each expression. Three times one is three. We then multiply the outer terms. Three times negative two 𝑖 is negative six 𝑖. The inner terms, negative 𝑖 times one is negative 𝑖. And then, negative 𝑖 times negative two 𝑖 is two 𝑖 squared. But of course, 𝑖 squared is negative one. So, this simplifies to three minus seven 𝑖 minus two, which is one minus seven 𝑖. And we could repeat that process for one plus two 𝑖 times one minus two 𝑖.

However, when we multiply a complex number by its conjugate, the result is the sum of the squares of the real and imaginary parts. So, that’s one plus four, which is equal to five. Now, let’s split this complex number into a fifth minus seven-fifths 𝑖. And we see that we need to solve the equation 𝑧 squared equals one-fifth minus seven-fifths 𝑖.

Now, it’s not particularly nice to take by the positive and negative square root of a fifth minus seven-fifths 𝑖. So, let’s say the solution to our equation is 𝑧 equals π‘Ž plus 𝑏𝑖. That’s just a complex number where π‘Ž and 𝑏 are real constants. Then, we know 𝑧 squared is π‘Ž plus 𝑏𝑖 all squared, which is π‘Ž squared minus 𝑏 squared plus two π‘Žπ‘π‘–. We can replace 𝑧 squared with this expression. And we find that π‘Ž squared minus 𝑏 squared plus two π‘Žπ‘π‘– must be equal to a fifth minus seven-fifths 𝑖.

And we can now equate the real and complex parts. On the left, the real part is a fifth. And on the right, it’s π‘Ž squared minus 𝑏 squared. So, a fifth equals π‘Ž squared minus 𝑏 squared. Then, the imaginary parts, on the left, we get negative seven-fifths. And on the right, we get two π‘Žπ‘. So, negative seven-fifths equals two π‘Žπ‘.

Let’s make 𝑏 the subject in the second equation. Dividing through by two π‘Ž, and we find 𝑏 is equal to negative seven over 10π‘Ž. We replace 𝑏 in our first equation with negative seven over 10π‘Ž, so we get a fifth equals π‘Ž squared minus negative seven over 10π‘Ž all squared. We square negative seven over 10π‘Ž to get 49 over 100π‘Ž squared. We then multiply through by π‘Ž squared and set this equal to zero. So, we get π‘Ž to the fourth power minus a fifth π‘Ž squared minus 49 over 100 equals zero.

Using whatever method we feel comfortable with, we can solve this equation. If we use the quadratic formula for example, instead of getting the solution for π‘Ž, we get the solution for π‘Ž squared. And we get one plus five root two over 10, or one minus five root two over 10. However, one minus five root two over 10 is negative. So, if we were to square root it, we get an imaginary number. And we know that π‘Ž itself needed to be a real constant.

So, π‘Ž is equal to the square root β€” in fact, the positive and negative square root β€” of one plus five root two over 10. That’s positive or negative 0.8983. And then, we replace π‘Ž with this number in our equation for 𝑏. And we get 𝑏 equals negative or positive 0.77917. And so, when we find the square root of 𝑧, that gives us both our positive and negative solutions. So, we have 𝑧 equals 0.898 minus 0.779𝑖 and 𝑧 equals negative 0.898 plus 0.779𝑖.

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