How do you divide (1+9i)/(-2+9i)?

Jul 27, 2016

$\frac{79}{85} - \frac{27}{85} i$.

Explanation:

Let $z = \frac{1 + 9 i}{- 2 + 9 i}$

Then usual Method to simplify this is to multiply both $N r .$ and $D r .$ by the conjugate of the complex no. of the $D r .$, i.e., to say,

$z = \frac{1 + 9 i}{- 2 + 9 i} \times \frac{- 2 - 9 i}{- 2 - 9 i}$

$= \frac{- 2 - 9 i - 18 i - 81 {i}^{2}}{- {2}^{2} - {\left(9 i\right)}^{2}}$

$= \frac{- 2 - 27 i - 81 \left(- 1\right)}{4 - 81 {i}^{2}}$

$= \frac{79 - 27 i}{85}$

$= \frac{79}{85} - \frac{27}{85} i$.

But, we can, as well, solve it as follows :-

Suppose that, $z = x + i y = \frac{1 + 9 i}{- 2 + 9 i}$

$\Rightarrow \left(x + i y\right) \left(- 2 + 9 i\right) = 1 + 9 i$

$\Rightarrow - 2 x + 9 i x - 2 i y + 9 {i}^{2} y = 1 + 9 i$

$\Rightarrow - 2 x + i \left(9 x - 2 y\right) + 9 \left(- 1\right) y = 1 + 9 i$

$\Rightarrow \left(- 2 x - 9 y\right) + i \left(9 x - 2 y\right) = 1 + 9 i$

Comparing the Real and Imaginary Parts of both sides, we get,

$\left(1\right) : 2 x + 9 y = - 1 , \mathmr{and} , \left(2\right) : 9 x - 2 y = 9$

$\therefore 2 \times \left(1\right) + 9 \times \left(2\right) \Rightarrow \left(4 x + 18 y\right) + \left(81 x - 18 y\right) = - 2 + 81$

$\Rightarrow 85 x = 79$ $\Rightarrow x = \frac{79}{85}$.

Using this with $\left(2\right)$, we have,

$2 y = 9 \left(x - 1\right) = 9 \left(\frac{79}{85} - 1\right) = 9 \left(- \frac{6}{85}\right)$

$\Rightarrow y = - \frac{9 \cdot 6}{85 \cdot 2} = - \frac{27}{85}$, giving,

$z = x + i y = \frac{79}{85} - \frac{27}{85} i$, as before!

Enjoy Maths!