Multiply $iln$ by $ \dfrac{a-bi}{a+bi} $ to get $ \dfrac{-bi^2ln+ailn}{bi+a} $.

Step 1: Write $ iln $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} iln \cdot \frac{a-bi}{a+bi} & \xlongequal{\text{Step 1}} \frac{iln}{\color{red}{1}} \cdot \frac{a-bi}{a+bi} \xlongequal{\text{Step 2}} \frac{ iln \cdot \left( a-bi \right) }{ 1 \cdot \left( a+bi \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ ailn-bi^2ln }{ a+bi } = \frac{-bi^2ln+ailn}{bi+a} \end{aligned} $$