Explanation

(a)
Here, $x = 4,\ \ \ \ y = - 3$
Let $x = r\cos{\theta,\ \ y = r\sin\theta} = - 3$
Now, $r = \sqrt{x^{2} + y^{2}} = \sqrt{16 + 9} = 5$
and $\theta = \tan^{- 1}\left( - \frac{3}{4} \right)$
now, let $R$ and $\phi$ be the magnitude and angle of resultant complex number.
$\therefore$ According to question.
$R = 3r$ and $\phi = \pi + \theta$
$\Rightarrow \ \ \ \ \ \ \ \phi = \pi + \tan^{- 1}\left( - \frac{3}{4} \right)$
$= \pi - \tan^{- 1}\left( \frac{3}{4} \right) = - \tan^{- 1}\left( \frac{3}{4} \right)$
$\therefore\ \ \ \ \cos{\phi = - \frac{4}{5},\ \ \ \ \sin{\phi = \frac{3}{5}\ \ \ \ \lbrack\because\ \phi\ lies\ in\ IInd\ quadrant\rbrack\ }}$
Hence, new complex number will be
$R\left( \cos{\phi + i\sin\phi} \right) = 3.5\ \left( - \frac{4}{5} + i\frac{3}{5} \right)$
$= \frac{3.5}{5}( - 4 + 3i) = - 12 + 9i$