Absolute stability of forward Euler with complex eigenvalues.

Question

Given the IVP : $$ y' = \lambda y, y(0) = y_0 $$

the region of absolute stability is given from $ |1 + h\lambda| < 1. $ where h is a fixed step size.

Let $\lambda = a + bi $ where $ a < 0 $ and $ b \neq 0 $

I have found an expression for h but I don't know if it is correct or if my approach was correct. Anyhow, here is what I did:

$ |1 + h\lambda| = |1 + hRe(\lambda) + hIm(\lambda)i| < 1$

Calculating the absolute value of a complex number gives:

$ \sqrt{(1 + hRe(\lambda))^2 + (hIm(\lambda))^2} < 1 $

This gives:

$ 1 + 2hRe(\lambda) + h^2(Re(\lambda))^2 + h^2(Im(\lambda))^2 < 1 $

Factorizing h:

$ h(h((Re(\lambda))^2 + (Im(\lambda))^2) + 2Re(\lambda)) < 0 $

Now I assume h can't be negative, which leaves me with the equation:

$ h((Re(\lambda))^2 + (Im(\lambda))^2) + 2Re(\lambda) < 0 $

Finally, solving for h gives:

$ h < \frac{-2Re(\lambda)}{(Re(\lambda))^2 + (Im(\lambda))^2} $

Is this answer correct? If not, then what is?