Melody

Yes I expect you are right.

I just solved it and got the silliest answer ever. (there were multiple places where a careless error was probably)

I got

\(a^2+b^3=\frac{324\pi^4-(4320-27\sqrt2)\pi^3+(12096+540\sqrt2)\pi^2+(15360-1456\sqrt2)\pi+(4096+8000\sqrt2))}{2048}\)

Somehow I do not think that is what they want    LOL

Are there any restrictions on a and b.  Like do they have to be real or positive or anything?

I will admit that this does not fall out as easily as I had hoped. I do not have an answer yet.

I have not finished this but this is how I would start.

\(x=\frac{a+5i}{1-2bi}\\ x=\frac{a+5i}{1-2bi}\times \frac{1+2bi}{1+2bi}\\ x=\frac{(a+5i)(1+2bi)}{(1-2bi)(1+2bi)}\\ etc\\ x \;\;also\;\;=\sqrt2+\frac{3\pi}{4}i\)

Equate coefficients and solve simultaneously.

Yes that is right.  (sorry about the confusion) 

\(\displaystyle \int_2^3(g'(x)-1)dx\\~\\ =\displaystyle \int_2^3(g'(x))dx+\displaystyle \int_2^3(-1)dx\\~\\ =\displaystyle \left[ g(x) \right]_2^3+\displaystyle \left[ -x \right]_2^3\)

you should be able to finish it from there.  Tell me what you get.

Yes you probably are,

Can you answer now or do you want me to keep going?

Actually the integral of  g'(x) is g(x) plus c  but since this is a definite integral you do not need to worry about the c. (it cancels out)

please just answer my question

Right so if you integrate g'(x) what do you get?

If you differentiate g(x)  with respect to x, what do you get?

Please answer, it is a simple question.