# Inequality of a complex number

## Homework Statement

Suppose that w is a complex number which is not both real and $\left\lfloor$w$\right\rfloor$$\geq$1 (the absolute value of w).
Verify that Re[(1-w$^{2}$)$^{1/2}$+iw]>0.

## The Attempt at a Solution

I attempted to solve this problem by dividing it into three cases; Im(w)>0, w$\in$(-1,1), and Im(w)<0. I could make the conclusion in the case of w$\in$(-1,1).
But, I don't have any idea how to approach in the cases of Im(w)>0 and Im(w)<0.
Could you give me a hint??

lanedance
Homework Helper
haven't worked it through fully but here's some stuff to get you going

rather than using cases, i'd consider the following reperesentations
$$|w|>1 \ \to w=re^{i\theta} \ , \ r>1$$
$$|w|>1 \ \to w=a+ib \ , \ a^2 + b^2 >1$$

can you convince yourself
$$Re\{(1-w^2)^{1/2}+iw\} = Re\{(1-w^2)^{1/2}\} +Re\{iw\}$$

an then you have
$$Re\{(1-w^2)^{1/2}\} +Re\{iw\}> 0$$

this is just a normal equality on the reals, so you do the normally allowed operations
$$Re\{(1-w^2)^{1/2}\} > -Re\{iw\}= Re\{-iw\}$$

then note
$$-Re\{iw\}= Re\{-iw\}= Im\{w\}$$