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Question Number 67992 by MJS last updated on 03/Sep/19

$$\left(1\right)z=a+b\mathrm{i}$$$$\left(2\right)z=r{\mathrm{e}}^{\mathrm{i}\theta }$$$$\mathrm{express}\mathrm{the}\mathrm{values}\mathrm{of}$$$$\left(\mathrm{a}\right)\mathrm{real}\left(z^z\right)\left[\mathrm{real}\mathrm{part}\right]$$$$\left(\mathrm{b}\right)\mathrm{imag}\left(z^z\right)\left[\mathrm{imaginary}\mathrm{part}\right]$$$$\left(\mathrm{c}\right)\mathrm{abs}\left(z^z\right)\left[\mathrm{absolute}\mathrm{value}\right]$$$$\left(\mathrm{d}\right)\arg \left(z^z\right)[\mathrm{argument}=\mathrm{angle}]$$

Answered by mr W last updated on 03/Sep/19

$$letA=z^z$$$$\ln A=z\ln z=(a+bi)(\ln r+i\theta )$$$$=(a\ln r-b\theta )+(a\theta +b\ln r)i$$$$A=e^{a\ln r-b\theta }{e}^{(a\theta +b\ln r)i}$$$$=e^{a\ln r-b\theta }\{\cos (a\theta +b\ln r)+i\sin \left(a\theta \times b\ln r\right)\}$$$$$$$$\mathrm{real}\left(z^z\right)=e^{a\ln r-b\theta }\cos (a\theta +b\ln r)$$$$imag\left(z^z\right)=e^{a\ln r-b\theta }\sin (a\theta +b\ln r)$$$$abs\left(z^z\right)=e^{a\ln r-b\theta }$$$$arg\left(z^z\right)=a\theta +b\ln r$$

Commented by MJS last updated on 03/Sep/19

$$\mathrm{thank}\mathrm{you}$$$$\mathrm{funny},\mathrm{I}\mathrm{did}\mathrm{it}\mathrm{exactly}\mathrm{the}\mathrm{same}\mathrm{way},\mathrm{had}\mathrm{been}$$$$\mathrm{unsure}\mathrm{if}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{right}...$$

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