Put two over negative five plus five 𝑖 in the form 𝑎 plus 𝑏𝑖.
In this question, we’re dividing a purely real number by a complex number. And so we recall that to perform these kinds of problems, we make sure our quotient is written as a fraction, and then we multiply both the numerator and the denominator by the conjugate of the denominator. Now, the conjugate of a complex number is simply found by changing the sign of the imaginary part. So, for a complex number 𝑧 in the form 𝑎 plus 𝑏𝑖, the conjugate, which is denoted 𝑧 bar or sometimes 𝑧 star, is 𝑎 minus 𝑏𝑖. And so, if we think about the denominator of the fraction we have, if we change the sign of the imaginary part, we get the conjugate to be negative five minus five 𝑖.
And so, our next job is going to be to multiply both the numerator and the denominator of our fraction by this complex number. Let’s begin by multiplying the denominator by negative five minus five 𝑖. We start by multiplying the first term in each expression. So, negative five times negative five is 25. We then multiply the outer terms. Negative five times negative five 𝑖 is positive 25𝑖. We multiply the inner terms, and we get negative 25𝑖. And finally, we multiply the last terms, and that gives us negative 25𝑖 squared.
Next, we notice that 25𝑖 minus 25𝑖 is zero. But we also know that 𝑖 squared is equal to negative one. So, this expression becomes 25 minus 25 times negative one. Well, negative 25 times negative one is positive 25. So we actually get a result of 50. The denominator of our fraction then is 50. The numerator is a little more straightforward. We simply multiply the two by negative five and negative five 𝑖. And that gives us negative 10 minus 10𝑖. So we see that our complex number is negative 10 minus 10𝑖 all over 50.
Now, in fact, we can simplify this even further. We divide each term by 10. And then we split the fraction up and we get negative one-fifth minus one-fifth 𝑖, which using decimals is negative 0.2 minus 0.2𝑖. And we’ve put our number in the form 𝑎 plus 𝑏𝑖. It’s negative 0.2 minus 0.2𝑖.