Electrical – Cascode Input Impedance

Ignore the capacitors, cable etc. and just consider the resistors. From probe tip to ground you have 9MΩ in series with 1MΩ - that's 10MΩ! You can verify this by measuring the resistance between probe tip and ground with a multimeter (on 20MΩ scale).

Of course this only applies to DC. At higher frequencies the impedances of the capacitors and cable become significant. At 50MHz the reactance of a 10pF capacitor is about 300Ω.

Small signal is all about change. Input impedance merely means the ratio of the change in input voltage at the point by some source and the change in current obtained by it. You could wiggle a current through the node too and measure the change in voltage at the node.

If you pass a small current di into the node (causing the total current to become I+di), since di = gm.dvgs (gm is transconductance), the change in voltage developed is 1/gm times di. Thus, the answer would be dv/di = 1/gm. The resistor appears as a resistor to ground (a small current change would occur through the resistor since other voltages are fixed giving the impression of resistance to the current source that caused the di) and thus zin = 1/gm || r.