### Instruction: Suppose that $a$ and $b$ are positive integers such that $(a+bi)^2 = 3+4i$. What is $a+bi$?### Response: I notice that the problem is asking for a complex number in the form of $a+bi$, where $a$ and $b$ are positive integers.
I also notice that the given equation $(a+bi)^2 = 3+4i$ is equivalent to $a^2 - b^2 + 2abi = 3+4i$.
To find $a$ and $b$, I need to match the real and imaginary parts of both sides of the equation.
That is, I need to solve the system of equations: $a^2 - b^2 = 3$ and $2ab = 4$.
The second equation, $2ab = 4$, implies that $ab = 2$.
So $a$ and $b$ must be factors of 2.
Since they are positive integers, the only possibilities are $a = 1$ and $b = 2$, or $a = 2$ and $b = 1$.
I plug in these possibilities into the first equation, $a^2 - b^2 = 3$, to see which one works.
If $a = 1$ and $b = 2$, then $a^2 - b^2 = 1 - 4 = -3$, which does not equal 3.
So this option is eliminated.
If $a = 2$ and $b = 1$, then $a^2 - b^2 = 2^2 - 1^2 = 4 - 1 = 3$, which does equal 3.
So this option is valid.
Therefore, I conclude that $a = 2$ and $b = 1$ are the values that satisfy the given equation.