11TH•Mathematics•Complex Numbers•Question 2

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

Express the following complex numbers in polar form:

\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)}

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(vii) 3+4i4+3i\dfrac{3+4i}{4+3i}4+3i3+4i​