How to rewrite $(9-i)/(3-i)$ in its standard form

Question

How do you rewrite $\frac{9-i}{3-i}$ in its standard form? At first, I thought that the expression itself was in standard form, but standard form would be $a+bi$, where $a$ and $b$ are constant values. I am struggling with this problem and I have tried a lot but with no proper results. I would be thankful if you helped me.

Answer 1

Your initial thought in a problem like this would be to simplify to find the expression in standard form. To do this, you multiply the numerator and denominator by $3+i$, so you get

$\frac{9-i}{3-i}*\frac{3+i}{3+i} = \frac{(9-i)(3+i)}{9-i^2} =\frac{27+6i-i^2}{9+1} = \frac{27+1+6i}{10}=\frac{28+6i}{10} = \frac{14}{5}+\frac{3}{5}i$

Therefore, $\frac{9-i}{3-i}$ in standard form would be equal to $\frac{14}{5}+\frac{3}{5}i$, from which you can also determine the values $a=\frac{14}{5}$ and $b=\frac{3}{5}$.

Answer 2

Multiply it by $\frac{3+i}{3+i}$.

In general, one multiplies $\frac{a+bi}{c+di}$ by $\frac{c-di}{c-di}$, because multiplication of a complex number with its conjugate gives you a real number, and you want a real number in your denominator.

The same idea is applied to square roots in fractions: multiply $\frac{a+\sqrt{b}}{c+\sqrt{d}}$ by $\frac{c-\sqrt{d}}{c-\sqrt{d}}$ to make the square root in the denominator disappear.

Answer 3

$$\frac {9-i}{3-i}=1+\frac {6}{3-i}=$$

$$1+\frac {6 (3+i)}{9+1}=$$ $$=1+\frac {3}{5}(3+i)=$$ $$\frac {14}{5}+\frac {3}{5}i$$