First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

(vii) $\dfrac{3+4i}{4+3i}$

\begin{aligned} \frac{3+4i}{4+3i} &=\frac{(3+4i)(4-3i)}{(4+3i)(4-3i)}\\ &=\frac{24+7i}{25}=\frac{24}{25}+\frac{7}{25}i \end{aligned}

r=\sqrt{\left(\frac{24}{25}\right)^2+\left(\frac{7}{25}\right)^2}=1

Since $x>0$ and $y>0$ :

\theta=\tan^{-1}\left(\frac{7}{24}\right)

\boxed{\frac{3+4i}{4+3i}=\cos\left(\tan^{-1}\frac{7}{24}\right)+i\sin\left(\tan^{-1}\frac{7}{24}\right)}