$$Log(-e i)$$
My try:
$$=\ln|0+(-e)i|+i[\arg (0+(-e i))+2\pi k]$$
$$=\ln|e i|+i(-\frac{\pi}{2}+2\pi k)$$
My attempt is correct?
$\endgroup$2 Answers
$\begingroup$$$...=\ln(|i||e|)$$
$$=\ln(|i|)+\ln(|e|)$$
$$=\ln(1)+\ln(|e|)$$
$$=0+1$$
$$\Longrightarrow \boxed{1+i(-\frac{\pi}{2}+2 \pi k)}$$
$\endgroup$ 1 $\begingroup$Correct but incomplete. The absolute value of ei is e, and the ln of that is 1.
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