Let $$O$$ be the origin and $$A$$ be the point $$z_1 = 1 + 2i$$. If $$B$$ is the point $$z_2$$, $$\text{Re}(z_2) < 0$$, such that $$OAB$$ is a right angled isosceles triangle with $$OB$$ as hypotenuse, then which of the following is NOT true?

We are given O (origin), A = $$z_1 = 1 + 2i$$, and B = $$z_2$$ with $$\text{Re}(z_2) < 0$$, such that OAB is a right-angled isosceles triangle with OB as hypotenuse.

We start by using the Pythagorean theorem and the isosceles condition. Since OB is the hypotenuse, the right angle is at A, and OA = AB.

By the Pythagorean theorem: $$|z_2 - z_1|^2 + |z_1|^2 = |z_2|^2$$

Let $$z_2 = a + bi$$:

$$(a-1)^2 + (b-2)^2 + 5 = a^2 + b^2$$

Therefore, $$-2a - 4b + 10 = 0 \implies a + 2b = 5 \quad \cdots(1)$$

Next, applying OA = AB gives $$|z_1|^2 = |z_2 - z_1|^2$$.

Hence, $$5 = (a-1)^2 + (b-2)^2 \quad \cdots(2)$$

Then, from (1) we have $$a = 5 - 2b$$. Substituting into (2) yields

$$(4 - 2b)^2 + (b - 2)^2 = 5$$

which can be rewritten as $$4(b-2)^2 + (b-2)^2 = 5$$

Hence, $$5(b-2)^2 = 5 \implies b = 3 \text{ or } b = 1$$

If $$b = 3$$ then $$a = -1$$, which satisfies $$\text{Re}(z_2) < 0$$, whereas $$b = 1$$ leads to $$a = 3$$ and is therefore rejected.

As a result, $$z_2 = -1 + 3i$$.

Finally, verifying each option shows that for Option A, $$\arg(z_2) = \arg(-1 + 3i) = \pi - \tan^{-1}3$$, which is TRUE.

Similarly, Option B gives $$z_1 - 2z_2 = (1+2i) - 2(-1+3i) = 3 - 4i$$ so that $$\arg(3 - 4i) = -\tan^{-1}\frac{4}{3}$$, which is TRUE.

Option C yields $$|z_2| = \sqrt{1 + 9} = \sqrt{10}$$, which is TRUE.

However, Option D produces $$2z_1 - z_2 = 2(1+2i) - (-1+3i) = 3 + i$$ and hence $$|3 + i| = \sqrt{10} \neq 5$$, making it NOT TRUE.

Therefore, the correct answer is Option D.