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Hi candlestick, I agreed with woodlouse's answer. Which solver gave you something different? I used this one:

https://www.emathhelp.net/calculators/algebra-2/solve-for-x-calculator/

or this if you want the steps:

https://www.emathhelp.net/calculators/algebra-1/quadratic-equation-calculator-solver/

The key number is +/- the square root of -1335, divided by 6

Hi, candlestick,

You have forgotten to adjust the quadratic equation for the 11/3 term on the right hand side. Deduct this from both sides of the equation to get x^2 + 7x + 48 1/3 = 0, then use a quadratic equation solver (or the formula we were taught in school) to get the two solutions already given.

Sorry, the quadratic should be x^2 + 7x + 49 1/3 = 0. Slip of the finger!

Thanks, I realized as you were posting that this is where I was going wrong. So I think I have my solution as to how the grid is to be filled, and no cell is going to be entirely blank.

Really don't like this reasonably business though, and I'm not happy about the way the letter values in brackets impacts on 14dn with regard to 38 and 41, as discussed earlier. This latter could and should have been fixed (unless there is some way of entering 38 41 that I'm not seeing).

I've just looked up Argand diagram and while all the individual words were familiar, the overall thing might as well have been written in Swahili. I shall content myself with completing the grid and maybe, when I feel I little stronger, take a look at some of the links offered. Many thanks to all the maths experts for the hints and tips but I think complex numbers and the like can stay firmly in my past.

What a nightmare that was!!! I now have a full grid and a headache - but I still need to fill in the diagram, which I think will be another hurdle yet to cross. As the better half is insisting I do some gardening, which isn't helping, I shall return to this later today, and probably tomorrow as well. Some of the clues, and the misprints, were in my opinion bordering on unfair, but just to the right side of it. Having "trigger" for 13ac proved rather distracting - but then again, I couldn't parse it properly anyway.

I assume "i" to the first decimal place is reasonable enough. If not, the hare moves into second place in the league of unreasonableness.

The solution to the equation gives the i part as plus or minus 6.5 which is the middle of the top and bottom cell in the third column. I agree with cockie that the Xs have to move there to give a correct Argand diagram. There is no need for the original Xs so I think there should be two blank cells at the end (hopefully!)

I would imagine that to be the case, Gitto. The combination of the thickness of the grid line, the thickness of someone's pen and the imprecision of handwriting all make it unlikely that the marker is going to distinguish between an X that is on the line (+/- 6) and one that is very slightly above or below for a more accurate representation. One tenth of a cell is .7mm unless you print a larger grid. I can't believe that level of accuracy is expected. If it is, then the marker will have to mark wrong those solutions where the X is not exactly equidistant between the two vertical grid lines.

Are you sure about that, unclued? The solution I obtained from three different sites gave +/- 6.0896...
That places it roughly on the line, or minutely above or below.