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Show by evaluating both sides that for $z_1 = i$ and $z_2 = -\sqrt 3 + i$,

$\log z_1z_2 \neq \log z_1 + \log z_2$.

Recall the definition: $$\log z = \log |z| + i\arg z$$

Attempt:

Left side:

$\log z_1z_2 = \log [i (-\sqrt 3 + i)] = \log[ i^2 - \sqrt 3i] = \log |- 1 - \sqrt 3 i| + i\arg(-1 - \sqrt 3i) = \log 2 + i(4\pi/3)$

and right side:

$\log z_1 + \log z_2 = \log i + \log (-\sqrt 3 + i) = {\log|i| + i\arg(i)} + {\log|-\sqrt 3 + i)|} + i\arg(-\sqrt 3 + i) = \log 1 + i (\pi/2) + \log 2 + i(5\pi/6) = \log 2 + i(\pi/2 + 5\pi/6) = \log 2 + i(4\pi/3) $.

But I get the same answer. Can someone please tell where the error is? Any feedback will really help. Thank you.

Community
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user384024
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  • Note that whether equality holds depends on which branch of $\log$ is used: Since $\arg i = \frac{\pi}{2}$ and $\arg (-\sqrt{3} + i) = \frac{4 \pi}{3}$, equality holds, e.g., for the (frequently used) branch whose argument takes values in $[0, 2 \pi)$. – Travis Willse Apr 10 '15 at 03:00

2 Answers2

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The $Arg(z)$ function spits out the principle argument, which is in the range $(-\pi,\pi]$. So your answer to the first one is wrong, it should be $\frac {-2\pi}3 i$ as the imaginary part

Alan
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  • is the second part fine? – user384024 Apr 10 '15 at 02:06
  • Yes, and this is the behavior that shows why the rule fails...adding two principle arguments often leads to a non principle one – Alan Apr 10 '15 at 02:07
  • how did you find $-2\pi/3$? as the correct angle? – user384024 Apr 10 '15 at 02:08
  • I subtracted $2\pi$ from your answer, to get it into the right interval. Remember that $2\pi$ is the period of the unit circle, if you are outside of that, add or subtract in incriments of $2\pi$ until you get to within $(-\pi,\pi]$ – Alan Apr 10 '15 at 02:10
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The principal-branch Arg is always between $-\pi$ and $\pi$, so $\text{Arg}(-1-i\sqrt{3})$ is not $4\pi/3$.

Robert Israel
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