Let [latex]\sqrt{1-z^2}=1[/latex] for z=0. Show that as z varies from 0 to p>1 along the real axis, [latex]\sqrt{1-z^2}[/latex] varies from 1 to [latex]-i \sqrt{p^2-1}[/latex].

Consider the case where z travels along path ABDEF, where BDE is a semi-circle as shown in Fig. 2-29. From this figure, we have 1 - z = 1 - x - iy = r cos θ - ir sin θ so that [latex]\sqrt{1-z^2}=\sqrt{(1-z)(1+z)}=\sqrt{r}(\cos \theta / 2-i \sin \theta / 2) \sqrt{2-r \cos \theta+i r \sin \theta}[/latex] Along AB: z = x, r = x -1, θ = 0 and [latex]\sqrt{1-z^2}=\sqrt{1-x} \sqrt{1+x}=\sqrt{1-x^2}[/latex]. Along EF: z = x, r = x -1, θ = π and [latex]\sqrt{1-z^2}--i \sqrt{x-1} \sqrt{x+1}--i \sqrt{x^2-1}[/latex]. Hence, as z varies from 0 [where x=0 ] to p [where x=p ], [latex]\sqrt{1-z^2}[/latex] varies from 1 to [latex]-i \sqrt{p^2-1}[/latex].

Question 2.43: Let √1-z^2=1 for z = 0. Show that as z varies from 0 to p &...

Schaum’s Outline of Complex Variables

Let $\sqrt{1-z^2}=1$ for z=0. Show that as z varies from 0 to p>1 along the real axis, $\sqrt{1-z^2}$ varies from 1 to $-i \sqrt{p^2-1}$ .

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Consider the case where z travels along path ABDEF, where BDE is a semi-circle as shown in Fig. 2-29. From this figure, we have

1 – z = 1 – x – iy = r cos θ – ir sin θ

so that $\sqrt{1-z^2}=\sqrt{(1-z)(1+z)}=\sqrt{r}(\cos \theta / 2-i \sin \theta / 2) \sqrt{2-r \cos \theta+i r \sin \theta}$

Along AB: z = x, r = x -1, θ = 0 and $\sqrt{1-z^2}=\sqrt{1-x} \sqrt{1+x}=\sqrt{1-x^2}$ .
Along EF: z = x, r = x -1, θ = π and $\sqrt{1-z^2}–i \sqrt{x-1} \sqrt{x+1}–i \sqrt{x^2-1}$ .
Hence, as z varies from 0 [where x=0 ] to p [where x=p ], $\sqrt{1-z^2}$ varies from 1 to $-i \sqrt{p^2-1}$ .