Seems pretty straight forward but absolute values have always given me headaches
$$\int_0^1 |1 -t + it|^2$$
Now usually I get roots and split up the intervals for when the function is greater or less than 0. But in this case I'm unsure what to do. The root of above is 0.5 + 0.5i. I tried doing
$$\int_0^1 |1 -t + it|^2 = \int_0^w (1 -t + it)^2 + \int_w^1 -(1 -t + it)^2 $$
where w = 0.5 + 0.5i. But this is wrong according to wolfram. I've also tried;
$$\int_0^1 |1 -t + it|^2 = |\int_0^1 (1 -t + it)^2 | $$
But I don't think I can do that because of the square? The answer doesn't match anyway. Guidance is appreciated.
Best Answer
You just have to remember that
$$|z| = \sqrt{ \mathcal{R}(z)^2 + \mathcal{I}(z)^2}$$
Hence
$$|1-t+it| = \sqrt{(1-t)^2+t^2}$$
Then
$$\int_0^1|1-t+it|^2 dt = \int_0^1 (1-t)^2+t^2 dt$$
and this is a really simple integral to calculate.