I know the answer for this is ii. I really don't remember seeing this problem in my readings though. Can someone please help me how to get √2 ?

 

Power transfer is maximized when load impedance is conjugately matched to source impedance: when power factor is 1. I would start this by transforming the source into a Thevenin source.
\(V_{S,Th.} = cos(t + 45^o) \cdot (1 - j/\omega) = cos(t + 45^o) \cdot |1 - j/\omega| \angle arctan(-1/1) = cos(t + 45^o) \cdot \sqrt{2} \angle -45^o = \sqrt{2} \angle 0^o V\)
\(Z_{S,Th.} = (1 - j/\omega) \Omega\)
Set load impedance for conjugate matching:
\(Z_L = Z_{S,Th.}*\)
\(1\Omega + Z_X = (1 - j/\omega)* = (1 + j/\omega) \Omega\)
Since \(\omega = 1\), \(Z_X = j/\omega = j\).
An inductor has impedance \(Z_{inductor} = j\omega L\), so L = 1 H.

If the answer truly is ii, I'd like to know where I went wrong.

 

Power transfer is maximized when load impedance is conjugately matched to source impedance: when power factor is 1. I would start this by transforming the source into a Thevenin source.
\(V_{S,Th.} = cos(t + 45^o) \cdot (1 - j/\omega) = cos(t + 45^o) \cdot |1 - j/\omega| \angle arctan(-1/1) = cos(t + 45^o) \cdot \sqrt{2} \angle -45^o = \sqrt{2} \angle 0^o V\)
\(Z_{S,Th.} = (1 - j/\omega) \Omega\)
Set load impedance for conjugate matching:
\(Z_L = Z_{S,Th.}*\)
\(1\Omega + Z_X = (1 - j/\omega)* = (1 + j/\omega) \Omega\)
Since \(\omega = 1\), \(Z_X = j/\omega = j\).
An inductor has impedance \(Z_{inductor} = j\omega L\), so L = 1 H.

If the answer truly is ii, I'd like to know where I went wrong.

First, isn't the 45° a red herring? 45° relative to what?
Second, I cheated and ran a simulation. Current through RL was maximum when L=1.

 

First, isn't the 45° a red herring? 45° relative to what?
Second, I cheated and ran a simulation. Current through RL was maximum when L=1.

Normally I would ignore this number as well, but you'll see that it cancels out with the -45° impedance branch, making subsequent steps (step?) slightly simpler. Good to know the math holds up...