Menu Close

Im-C-0-1-2-z-dz-pi-2-




Question Number 136682 by snipers237 last updated on 24/Mar/21
 Im(∫_(C^+ (0,(1/2))) z^− dz )= (π/2)
$$\:{Im}\left(\int_{{C}^{+} \left(\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)} \overset{−} {{z}dz}\:\right)=\:\frac{\pi}{\mathrm{2}}\:\: \\ $$
Answered by Olaf last updated on 24/Mar/21
Ω = Im∫_(C^+ (0,(1/2))) z^− dz  Ω = Im∫_0 ^(2π) (1/2)e^(−iθ) d((1/2)e^(iθ) )  Ω = (1/4)Im∫_0 ^(2π) idθ  Ω = (1/4)Im(2πi) = ((2π)/4) = (π/2)
$$\Omega\:=\:\mathrm{Im}\int_{\mathrm{C}^{+} \left(\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)} \overset{−} {{z}dz} \\ $$$$\Omega\:=\:\mathrm{Im}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{2}}{e}^{−{i}\theta} {d}\left(\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} \right) \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{Im}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {id}\theta \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{Im}\left(\mathrm{2}\pi{i}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{4}}\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *