(Note: $i$ refers to the imaginary unit)
I am trying to find the values of $\lambda$ for which $\alpha =i$ is a root of the quadratic:
$$z^2+\lambda z-4$$
Using the quadratic formula I got:
$$z=\frac{-\lambda \pm \sqrt{\lambda^2+16}}{2}$$
(Note: I believe that $\alpha$ refers to the root that is equal to $\frac{-b \textbf{+} \sqrt{b^2-4ac}}{2a}$, but I could be mistaken)
Therefore, to find the value of $\lambda$ for which $\alpha =i$, I have to solve the equation:
$$ i=\frac{-\lambda + \sqrt{\lambda^2+16}}{2}$$
Here is an outline of the steps I took:
Multiply both sides by $2$ and add $\lambda$ to both sides:
$$2i+\lambda=\sqrt{\lambda^2 +16}$$
Square both sides:
$$4i^2 +4\lambda i +\lambda^2 =\lambda^2 +16$$
Subtract $\lambda^2$ from both sides then solve for $\lambda$:
$$\lambda=-5i$$
However, $\lambda=-5i$ does not solve the equation ($ i \neq 4i$ ), which is very confusing. Where did I go wrong?
P.S: I did find that $ \lambda=-5i $ was a solution to the other root, $i=\frac{-\lambda – \sqrt{\lambda^2+16}}{2}$.
Best Answer
Each non-zero real or complex number $w$ has exactly two square roots $r_1, r_2$ which differ by sign ($r_2 = - r_1$). If $w$ is a positive real number, then both $r_1, r_2$ are real and the standard convention is to write $\sqrt w$ for the positive square root. In all other cases there is no universally accepted interpretation of $\sqrt w$. If $w$ is a negative real number, then one may of course use the definition $\sqrt w = i \sqrt {\lvert w \rvert}$ (as you do in your calculation). This is a good and reasonable choice for $\sqrt w$, but it is only one of two possible choices. For $w \notin \mathbb R$ it is even more arbitrary to define a unique value of $\sqrt w$.
You corrrectly state that the quadratic equation $$z^2 + \lambda z - 4 = 0 \tag{1}$$ has the two solutions $$z=\frac{-\lambda \pm \sqrt{\lambda^2+16}}{2} \tag{2}$$ As explained above, this formula is based on a specific interpretation of the expression $\sqrt{\lambda^2+16}$; it involves a convention concerning the symbol $\sqrt{\phantom x }$. Perhaps it would be a more neutral way to say that the solutions of $(1)$ have the form $$z=\frac{-\lambda + r}{2} \tag{3}$$ where $r$ is any square root of $\lambda^2+16$, i.e. any solution of $$r^2 = \lambda^2+16 . \tag{4}$$
If you want to have $i$ as a solution of $(1)$, you have to determine $\lambda \in \mathbb C$ and $r$ with $r^2 = \lambda^2+16$ such that $$i=\frac{-\lambda + r}{2} \tag{5}$$ You write right from the start $r = \sqrt{\lambda^2+16}$, but this is inadequate if you use a fixed convention for the symbol $\sqrt{\phantom x }$. In fact, you must be aware that only one of the two solutions of $(4)$ will satisfy $(5)$ and the other produces a value $ \ne i$. Can you be sure that a fixed square root convention produces the appropriate value for $r$? No, you can't.
Your calculation correctly gives $\lambda = -5i$ and thus $\lambda^2+16 = -9$. The above convention for square roots of negative real numbers produces then $r = \sqrt{-9} = 3i$ which does not satisfy $(5)$. The other solution of $(4)$ is $r = -3i$ and this is the adequate one which satisfies $(5)$.
Update:
A simpler approach is this. You want to find $\lambda$ such that $i$ is a root of $(1)$. Let $\rho$ be the second root of $(1)$. Then $$(z-i)(z-\rho) = z^2 + \lambda z - 4$$ which gives $i\rho = -4$ and $-(i + \rho) = \lambda$. The first equation implies $\rho = 4i$ and the second implies then $\lambda = -5i$.