# How do you simplify -9-9i \div -1-8i?

How do you simplify $-9-9i÷-1-8i$?
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mauricio0815sh
$-9-9i÷-1-8i=\frac{81}{65}-\frac{63}{65}i$
Explanation:
$-9-9i÷-1-8i=\frac{-9-9i}{-1-8i}$
$=\frac{-9-9i}{-1-8i}\cdot \frac{-1+8i}{-1+8i}$
$\frac{9+9i-72i+72}{1+64}$
$=\frac{81-63i}{65}$
###### Not exactly what you’re looking for?
Carl Swisher
Solution:
$-9-9i÷-1-8i$ can be written as $\frac{-9-9i}{-1-8i}$
To simplify we need to multiply numerator and denominator by the complex conjugate of the denominator.
Complex conjugate of a number
hence in above case one needs to multiply by $-1+8i$
Hence $\frac{-9-9i}{-1-8i}$
$=\frac{\left(-9-9i\right)×\left(-1+8i\right)}{\left(-1-8i\right)×\left(-1+8i\right)}$
$=\frac{9-72i+9i-72{i}^{2}}{{\left(-1\right)}^{2}-{\left(8i\right)}^{2}}$
$=\frac{9-72i+9i-72\left(-1\right)}{1-64\left(-1\right)}$
$=\frac{9-72i+9i+72}{1+64}$
$=\frac{81-63i}{65}$
$=\frac{81}{65}-\frac{63}{65}i$