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This problem is giving me quite a bit of grief. Heres my issue I do the first problem step by step and get the correct solution. I then try the second problem step by step and get the wrong answer. Specifically I get them to be equal again. Can someone explain why the process taken in question 1 does not work in 2. Thanks for the help. textbook solution

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    $\begingroup$ Please, write out the question instead of referring us to an image. $\endgroup$ Sep 17, 2017 at 23:24

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The expanation for the difference between Part (i) and Part (ii) is that in the complex numbers $e^u=e^v$ does $not$ imply $e^{u/5}=e^{v/5}$.

Note that $(\sqrt 3\; -i)^{1/5}$ is not a number but a set of $5$ numbers.

In Part (i), "$z$ is a log of $(\sqrt 3\;-i)^{1/5}\;$" iff $(e^z)^5=\sqrt 3 -i$..... while "$z$ is $1/5$ of a log of $\sqrt 3\;-i $" iff "$5z$ is a log of $\sqrt 3\;-i\;$" iff $e^{5z}=\sqrt 3\;-i.$ (... & observe that $(e^z)^5=e^{5z}.)$

In Part (ii), "$z$ is a log of $(\sqrt 3\;-i)^5\;$" iff $e^z=(\sqrt 3\;-i)^5,$ but $e^z=(\sqrt 3\;-i)^5$ does $not$ imply that $e^{z/5}=\sqrt 3\;-i$.... while "$z$ is $5$ times a log of $\sqrt 3\; -i\;$" iff $e^{z/5}=\sqrt 3\;-i.$

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  • $\begingroup$ Thank you that was a great explanation! $\endgroup$
    – TimHorton
    Sep 18, 2017 at 17:47

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