Multiplying two fractions with complex numbers

Question

I'm doing $$ \frac{6-7i}{1+i}\cdot\frac{1+i}{1+i}, $$ and I'm getting the correct value for the numerator (namely, $-1-13i$), but based on the problem answer, I need for the denominator to become $2$. However, $(1+i)^2\neq 2$. How can I write the denominator as a real number?

Answer 1

It appears that you are trying to write $$ \frac{6-7i}{1+i} $$ in the form $a+bi$ for some $a,b\in\mathbb R$. We can multiply this fraction by $1 = \frac{1-i}{1-i}$ in a clever way: $$ \frac{6-7i}{1+i}\cdot\frac{1-i}{1-i} = \frac{(6-7i)(1-i)}{(1+i)(1-i)}. $$ The numerator works out to $$ (6-7i)(1-i) = 6-6i-7i+7i^2 = -1-13i, $$ while the denominator works out to $$ (1+i)(1-i) = 1+i-i-i^2 = 1-(-1) = 2. $$ In this problem, you want to get rid of the $i$ in the denominator; in other words, given $a+bi$ you want to find $c+di$ such that $(a+bi)(c+di)$ is just a real number (no $i$ term). This is easily accomplished by setting $$c+di = \overline{a+bi} = a-bi,$$ the complex conjugate of $a+bi$. Then $$ (a+bi)(a-bi) = a^2+bi-bi-b^2i^2 = a^2+b^2. $$ However, if you were to take $a+bi$ and compute $(a+bi)^2$, like you were trying to do, then you would obtain $$ (a+bi)(a+bi) = (a^2-b^2)+2abi. $$ But then $2ab\neq 0$ unless $a=0$ or $b=0$. So unless you're lucky, doing $$ \frac{a+bi}{c+di}\cdot\frac{c+di}{c+di} $$ won't work so well to simplify that denominator.