$\frac{10+5\mathrm{i}}{3+4\mathrm{i}}$

Project ID: 
3000000004
Question: 

Romeo had to find the imaginary part of a complex number $\frac{10+5\mathrm{i}}{3+4\mathrm{i}}$.

In which step of his solution did Romeo make a mistake?

(The step number is above the equality sign.)

$$ \begin{aligned} \frac{10+5\mathrm{i}}{3+4\mathrm{i}} &\stackrel{(1)}= \frac{(10+5\mathrm{i})(3+4\mathrm i)}{9+16}= \cr &\stackrel{(2)}= \frac{30+40\mathrm i + 15\mathrm i +20\mathrm i^2}{25}=\cr &\stackrel{(3)}= \frac{10+55\mathrm i}{25} =\cr&\stackrel{(4)}= \frac 25 + \frac{11}5\mathrm i \stackrel{(5)}\implies \end{aligned} $$

The imaginary part of this complex number $\frac 25 + \frac{11}5 \mathrm i$ is $\frac{11}{5}$.

Answer 1: 

In step (1). Romeo multiplied the numerator by $3+4\mathrm i$ and the denominator by $3-4\mathrm i$.

Answer 2: 

In step (2). The expression simplifies to $\frac{30-40-15+20\mathrm i^2}{25}$.

Answer 3: 

In step (3). The expression simplifies to $\frac{50+55\mathrm i}{25}$.

Answer 4: 

In step (5). The imaginary part of the complex number $\frac 25 + \frac{11}5 \mathrm i$ is $\frac {11}5 \mathrm i$.

Correct Answer: 
Answer 1
Hint: 

$$ \begin{aligned} \frac{10+5\mathrm{i}}{3+4\mathrm{i}} &\stackrel{(1)}= \frac{(10+5\mathrm{i})(3-4\mathrm i)}{9+16} =\cr &\stackrel{(2)}= \frac{30-40\mathrm i + 15\mathrm i -20\mathrm i^2}{25}=\cr &\stackrel{(3)}= \frac{50-25\mathrm i}{25} =\cr&\stackrel{(4)}= 2 -\mathrm i\stackrel{(5)}\implies \end{aligned} $$

The imaginary part of this complex number is $-1$.