If a^2 + b^2 = 1, is (a + b) = 1? (1) ab = 0 (2) b = 0

From the task description we have that \(a^2 + b^2 = 1\). Note that in the task it is not specified that both a and b are integers. Since then, it can be possible that \(a =\frac{\sqrt{3}}{2}\) and \(b=\frac{1}{2}\). In this case \(a^2 + b^2 = (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{3}{4} = 1\)
Even though both \(a^2\) and \(b^2\) are positive, it is not necessarily that \(a\) and \(b\) are also positive. It is possible that \(a = 0\) and \(b = -1\): \(a^2 + b^2 = (0)^2 + (-1)^2 = 0 + 1= 1\)
Let us look closer to each statement.

Statement 1:
1) \(ab = 0\)
Since \(ab = 0\), we can conclude that either \(a\) or \(b\) is equal to \(0\). However, if we assume that \(b = 0\), \(a\) can be both \(1\) and \(-1\). In case \(a = -1\), \(a + b = (-1) + 0 = -1\) and in case \(a = 1\), \(a + b = 1 + 0 = 1\).
The statement is clearly insufficient.

Statement 2:
2) \(b = 0\)
This statement states the same thing that the statement 1 and thus, it is also insufficient.

Both statements 1 and 2 together state the same thing and for this reason both of them are insufficient.

Answer is E.