Which complex number is NOT in standard form? $$ 2 i $$ $$ 2-i^{2} $$ $$ 3+i $$

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To determine which complex number is **not** in standard form, let's first recall that the standard form of a complex number is:

$$
a + b i
$$

where:
- $$a$$ is the real part,
- $$b$$ is the imaginary coefficient,
- and $$i$$ is the imaginary unit ($$i^2 = -1$$).

Now, let's analyze each of the given complex numbers:

1. **$$2i$$**
   - This can be written as $$0 + 2i$$, which fits the standard form with $$a = 0$$ and $$b = 2$$.
  
2. **$$2 - i^2$$**
   - Simplify the expression using $$i^2 = -1$$:
     $$
     2 - i^2 = 2 - (-1) = 2 + 1 = 3
     $$
   - After simplification, it becomes $$3$$, which can be written as $$3 + 0i$$, fitting the standard form. However, **before simplification**, it involves $$i^2$$ rather than just $$i$$, meaning it's not initially presented in the standard form $$a + bi$$.
  
3. **$$3 + i$$**
   - This is already in the standard form with $$a = 3$$ and $$b = 1$$.

**Conclusion:**
While both $$2i$$ and $$3 + i$$ are in standard form, $$2 - i^2$$ is **not** initially presented in standard form because it includes $$i^2$$ instead of just $$i$$.

**Answer:** $$2 - i^{2}$$ is not written in standard form.
$$2 - i^{2}$$ is not in standard form.

Which complex number is NOT in standard form? \( 2 i \) \( 2-i^{2} \) \( 3+i \)

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