### Instruction: Simplify $\dfrac{3+4i}{1+2i}$. Your answer should be of the form $a+bi$, where $a$ and $b$ are both real numbers and written as improper fractions (if necessary).### Response: To simplify a complex fraction like this, I need to eliminate the denominator by multiplying both the numerator and denominator by its complex conjugate.
The complex conjugate of a complex number is the same number but with the opposite sign of the imaginary part.
So, the complex conjugate of $1+2i$ is $1-2i$.
Therefore, I multiply both the numerator and denominator by $1-2i$ and use the distributive property and the fact that $i^2=-1$ to simplify.
This gives me $\dfrac{(3+4i)(1-2i)}{(1+2i)(1-2i)}=\dfrac{3-6i+4i-8i^2}{1-2i+2i-4i^2}=\dfrac{3-2i+8}{1+4}=\dfrac{11-2i}{5}$.
I can leave the answer like this, or I can split it into two fractions: $\dfrac{11-2i}{5}=\dfrac{11}{5}-\dfrac{2}{5}i$.