Inequality of a complex number

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Homework Statement



Suppose that w is a complex number which is not both real and [itex]\left\lfloor[/itex]w[itex]\right\rfloor[/itex][itex]\geq[/itex]1 (the absolute value of w).
Verify that Re[(1-w[itex]^{2}[/itex])[itex]^{1/2}[/itex]+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[itex]\in[/itex](-1,1), and Im(w)<0. I could make the conclusion in the case of w[itex]\in[/itex](-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??
 

Answers and Replies

  • #2
lanedance
Homework Helper
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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
[tex] |w|>1 \ \to w=re^{i\theta} \ , \ r>1[/tex]
[tex] |w|>1 \ \to w=a+ib \ , \ a^2 + b^2 >1[/tex]

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

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

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

then note
[tex] -Re\{iw\}= Re\{-iw\}= Im\{w\}[/tex]
 

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