Inequality of a complex number

[gotjrgkr]

Homework Statement

Suppose that w is a complex number which is not both real and \left\lfloorw\right\rfloor\geq1 (the absolute value of w).
Verify that Re[(1-w^{2})^{1/2}+iw]>0.

Homework Equations

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]

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|&gt;1 \ \to w=re^{i\theta} \ , \ r&gt;1
|w|&gt;1 \ \to w=a+ib \ , \ a^2 + b^2 &gt;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\}&gt; 0

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

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