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

(1) z=a+bi (2) z=re^(iθ) express the values of (a) real (z^z ) [real part] (b) imag (z^z ) [imaginary part] (c) abs (z^z ) [absolute value] (d) arg (z^z ) [argument = angle]

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

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

let A=z^z ln A=z ln z=(a+bi)(ln r+iθ) =(aln r−bθ)+(aθ+bln r)i A=e^(aln r−bθ) e^((aθ+bln r)i) =e^(aln r−bθ) {cos (aθ+bln r)+i sin (aθ×bln r)} real (z^z )=e^(aln r−bθ) cos (aθ+bln r) imag (z^z )=e^(aln r−bθ) sin (aθ+bln r) abs (z^z )=e^(aln r−bθ) arg (z^z )=aθ+bln r

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

Commented by MJS last updated on 03/Sep/19

thank you funny, I did it exactly the same way, had been unsure if I′m right...

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

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