How could I rewrite $\dfrac{6 + 4 {i}} { -9 - 4 {i}}$ in a+bi form? Peace to all. When I solve the problem I get $\dfrac{70} { 97}$ - $\dfrac{12{i}} {97}$ and it's the wrong answer. How exactly do you go about solving this problem?
This is my work: I received that answer by multiplying both the numerator and denominator by the conjugate partner of "-9 - 4i" which is "-9 + 4i".
$\dfrac{-54 + 24 {i} - 36{i} +16i^2} { 81 - 36 {i} + 36 {i} - 16i^2}$
Combining like terms: $\dfrac{-54 + 24 {i} - 36{i} +16(-1)} { 81 - 36 {i} + 36 {i} - 16(-1)}$ = $\dfrac{-70 - 12i} { 81 + 16}$ = $\dfrac{-70} { 97}$ - $\dfrac{12{i}} {97}$

Another approach that can be taken is through the application of the definition of division.  If $ \ \frac{6 + 4 i}{ -9 - 4 i} \ = \ a + bi \ \ , $ then
$$ 6 \ + \ 4i \ \ = \ \ (-9 \ - \ 4i)·(a \ + \ bi) \ \ = \ \ -9a \ - \ 4ai \ - \ 9bi \ - \ 4b·i^2 \ \ . $$
Equating the real and imaginary parts produces the pair of equations $ \ -9a + 4b \ = \ 6 \ \ , \ \ -4a - 9b \ = \ 4 \ \ . $  Solving this system by multiplication and addition
[first array: multiply first equation by 9 , second by 4 ; second array: multiply first equation by (-4) , second by 9] yields
$$ \begin{array}{ccc} -81a & + \ 36b & \ = \ 54 \\ -16a &   - \ 36b & \ = \ 16 \end{array} \ \ \ , \ \ \  \begin{array}{ccc} 36a & - \ 16b & \ = \ -24 \\ -36a & \ - \ 81b & \ = \ 36 \end{array} $$
[then add equations together]
$$ \Rightarrow \ \ -97a \ \ = \ \ 70 \ \ \ , \ \ \ -97b \ \ = \ \ 12 \ \ . $$
Hence, the ratio is $ \ \large{\frac{6  \ + \  4 i}{ -9 \ - \ 4 i}} \ = \ \normalsize{-\frac{70}{97}   - \frac{12}{97}i} \ \ , $
in agreement with the result from other posters.
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The multiplication of the numerator and denominator of a ratio of complex numbers by the complex conjugate in order to "rationalize" the denominator is, of course, the conventional method of simplification.  It can be "reduced" to a formula (although the calculations are usually simple enough that the formula isn't generally taught):
$$ \frac{a \ + \ bi}{c \ + \ di} \ · \ \frac{c \ - \ di}{c \ - \ di}  \ \ = \ \ \frac{ac \ + \ bd}{c^2 \ + \ d^2} \ + \ i·\left(\frac{bc \ - \ ad}{c^2 \ + \ d^2} \right) \ \ ;  $$
for this problem, we then obtain
$$ \frac{6·(-9) \ + \ 4·(-4)}{(-9)^2 \ + \ (-4)^2} \ + \ i·\left(\frac{4·(-9) \ - \ 6·(-4)}{(-9)^2 \ + \ (-4)^2} \right) $$ $$ = \ \ \frac{(-54) \ + \ (-16)}{81 \ + \ 16} \ + \ i·\left(\frac{(-36) \ - \ (-24)}{81 \ + \ 16} \right) \ \ . $$
Because the method of solving a system of two linear equations is equivalent, there is a way of representing complex numbers as $ \ 2 \times 2 \ $ matrices, $ \ a + bi \ \rightarrow \ \left[\begin{array}{cc}  a &  b  \\ -b &   a  \end{array} \right] \ \ . $  [I mention this approach since many "algebra/pre-calculus" course at least introduce matrices.]  Multiplying two complex numbers then gives the same result as multiplying two such matrices and division involves multiplying the matrix representing the numerator by the inverse of the matrix representing the denominator.  For this problem,
$$ 6 + 4i \ \rightarrow \ \left[\begin{array}{cc}  6 &  4  \\ -4 &   6  \end{array} \right] \ \ , \ \ -9 - 4i \ \rightarrow \ \left[\begin{array}{cc}  -9 &  -4  \\ 4 &   -9  \end{array} \right] $$ $$ \Rightarrow \ \ \frac{1}{-9 - 4i} \ \ = \ \ \left[\begin{array}{cc}  -9 &  -4  \\ 4 &   -9  \end{array} \right]^{-1} \ \ = \ \ \frac{1}{(-9)·(-9) \ - \ 4·(-4)} \left[\begin{array}{cc}  -9 &  4  \\ -4 &   -9  \end{array} \right] \ \ = \ \ \frac{1}{97} \left[\begin{array}{cc}  -9 &  4  \\ -4 &   -9  \end{array} \right] \ \ ; $$
$$ \frac{6  \ + \  4 i}{ -9 \ - \ 4 i} \ \ \rightarrow \ \ \frac{1}{97} \left[\begin{array}{cc}  -9 &  4  \\ -4 &   -9  \end{array} \right] \ \left[\begin{array}{cc}  6 &  4  \\ -4 &   6  \end{array} \right] \ \ = \ \ \frac{1}{97} \left[\begin{array}{cc}  -54 - 16 &  -36+24  \\ -24 + 36 &   -16-54  \end{array} \right]  $$
$$ = \ \ \frac{1}{97} \left[\begin{array}{cc}  -70 &  -12  \\ 12 &   -70  \end{array} \right] \ \ , $$
which represents the complex number $ \ -\frac{70}{97}   - \frac{12}{97}i \ \ . $
