If z=1+2i is a root of the equation z^2 = az+ b find the values of a and b
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If you'd been given, say, [imath]x^2 = ax + b[/imath] and had been told that [imath]x = 2[/imath] is a solution, what would you have done? When you solved quadratics in the past, how would you get complex-valued solutions? (Hint: Quadratic Formula)
Dr.Peterson
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Is it assumed that a and b are real numbers, or can they be imaginary? I don't think this can be solved in the latter case. In the former case, what other number is a root? What factors would that imply?
Standard form for complex number z is a+bi.
If you have confusion about why a=1 and b=2 do not work, then you should be forgiven because they have written this exercise so that a and b mean different values in the equation than they do in the given root. Parameters should not change value within a single exercise, so they must want people to ignore standard form for z.
Dr. Peterson and staple posted some ideas. I think of another. Here is example.
z^2 = a*z+b
2-7i = root (given in my example)
Substitute given root for z in equation and then simplify each side.
(2-7i)^2 = a(2-7i)+b
-45-28i = 2a+b-(7a)i
Imaginary number on each side must be same.
-(28)i = -(7a)i
28 = 7a
a=4
Real number on each side must be same.
-45 = 2a+b
-45 = 2(4)+b
b=-53
Dr.Peterson
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No, mention of z does not implicitly define a and b as its parts; you can define any variables you want for those parts, such as x and y. There is nothing wrong in using a and b for something else in the problem. They did not already have definitions.
In your work, you implicitly assumed that a and b in the problem are real numbers; that is true of the real and imaginary parts, but not necessarily true of variables in a problem in the context of complex numbers.
If it were stated or implied that a and b are real numbers, as I said they should be in order for the problem to be solvable, then your approach would be valid.
I also think so. In my idea the equation is basic quadratic, the variable is z and the unknown coefficients (a and b) are real numbers.
pka
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Lets be sure of the question. Given that [imath]1+2i[/imath] is a root of [imath]z^2=az+b[/imath][imath]~,~\{a,b\} \subseteq \Re[/imath] THEN
Using the root [imath]\left\{ \begin{gathered} {(1 + 2i)^2} = a(1 + 2i) + b \\ 1 + 4i - 4 = a + 2ai + b \\ - 3 + 4i = ai + 2ai + b \\ \end{gathered} \right.[/imath]
Now gather terms setting equal to zero; the real parts equal & imaginary parts equal.
[imath]\left\{ \begin{gathered} \left( { - a - 3 - b} \right) + \left( {4 - 2a} \right)i = 0 \\ \left( { - a - 3 - b} \right) = 0\;\& \;\;(4 - 2a) = 0 \\ a = 2\;\& \;b = - 5 \\ \end{gathered} \right.[/imath]
[imath][/imath][imath][/imath]