So, what requires more thrust? A climb with 100fpm residual climb or a bank angle of 39 degrees and what is indicated by the T behind max altitude in the FMS: the ability to maintain airspeed or the ability to sustain a 100fpm climb.
I cannot answer that. However:
Rate of climb = excess power divided by weight
Excess power = excess thrust x velocity
Assume additional drag (and therefore thrust) for turn varies as square of the coefficient of lift
Assume flight at min drag speed to simplify
39 degrees AoB is a load factor of 1.28. Therefore the lift coefficient increase is 1.28 x whatever it was. 1.28 squared is 1.64. In other words, the induced drag [as opposed to total drag] is 64% greater than straight and level.
Total drag is lift dependant plus zero lift drag. At min drag speed straight and level that is simply 2 times the lift dependant drag, whilst at 39 deg AoB the total drag is now 2.64 times the lift dependant drag. In other words, drag and therefore thrust is 32% higher than straight and level. That deals with the turn thrust.
For climb, if you know your speed you can easily work out your excess thrust at 100 fpm:
ROC = (Excess thrust x velocity)/weight. Rearranging:
Excess thrust = (ROC x weight)/velocity
Now given flight at Vmd of say 100kts and ROC of 100 fpm (which is 1 knot):
Excess thrust = (1 x weight)/100
Excess thrust = weight/100
...eg @ 10000lb weight you would need 100lbs excess thrust to acheive 100 fpm.
...I'll just put that one out there for review. However, I suspect that if you factor in all variables in detail for high speed and high altitude flight you will get a more relevant answer than my crude theory would lead you to.