I am trying to solve this exercise.
Consider the following initial value problem $$ y^{\prime} (t) = \lambda y (t) \, \text{for} \, t \in [0; T] ; y (0) = y_0; $$ with $T > 0, y_0 \in \mathbb{R} $ and $ \lambda \in \mathbb{C}$. Consider the explicit Euler method. Is it possible to specify a boundry for the stepsize $h > 0$ for $$ a) \lambda = 3 + 3i; \, b) \lambda = -3 + 3i \, \text{where} \, i^2 = -1; $$ so that the method outputs a stable numerical solution?
I have tried to solve this exercise over the stability function of the Runge Kutta Method. It is known that $$ R(z) = 1+z$$ with $z = \lambda h$. And it is also known that $\vert Re(z) \vert < 1$ sould hold, if we want a stable method.
Thus, $$ \vert Re(z) \vert = \vert 1+z \vert = \vert 1+ \lambda h \vert \leq 1 + h \vert \lambda \vert \overset{a)}{=} 1+ h \vert 3+3i \vert = 1 + h \sqrt{18} \overset{!}{<} 1 \Rightarrow h<0$$
That is not possible because $h>0$
So, how can I handle this exercise? Or do I make a mistake?
I think you should not use the triangle inequality. In fact, for $\lambda=3+3i$, $$ |1+\lambda h|=|1+(3+3i)h|=|(1+3h)+3hi|=\sqrt{(1+3h)^2+(3h)^2}>1$$ for $h>0$. It means that the method is unstable for $\lambda=3+3i$. For $\lambda=-3+3i$, $$ |1+\lambda h|=|1+(-3+3i)h|=|(1-3h)+3hi|=\sqrt{(1-3h)^2+(3h)^2}<1$$ gives $$ 18h^2-6h<0 $$ whose solution is $0<h<\frac13$. Thus if $0<h<\frac13$, the method is stable for $\lambda=-3+3i$.