How to find the absolute value of this complex number: $\frac{-4-6i}{17+i}$
I know that, in general, $|a+bi|=\sqrt{a^2+b^2}$, however, I don't know how to make $\frac{-4-6i}{17+i}$ into the form of $a+bi$.
$\endgroup$ 12 Answers
$\begingroup$Notice:
$$\left|\frac{z}{s}\right|=\frac{\left|z\right|}{\left|s\right|}$$
So:
$$\left|\frac{-4-6i}{17+i}\right|=\frac{\left|-4-6i\right|}{\left|17+i\right|}=\frac{\sqrt{(-4)^2+(-6)^2}}{\sqrt{(17)^2+(1)^2}}=\frac{\sqrt{52}}{\sqrt{290}}=\sqrt{\frac{26}{145}}=\frac{\sqrt{3770}}{145}$$
$\endgroup$ $\begingroup$The usual trick is to multiply and divide the denominator by the conjugate of the denominator. Since $$z \cdot\bar{z} = |z|^2$$ This allows us make the denominator a real number.
$$ \frac{-4 - 6i}{17 + i} = \frac{(-4 - 6i) \times (17 - i)}{(17 + i) \times (17 - i)} = \\ \frac{-74 - 98i}{17^2 - i^2} = \frac{-74 - 98i}{290} = \\ \frac{-37}{145} - \frac{49i}{145} $$
From here on out, the absolute value is straightforward
$\endgroup$ 0
