Logs Of Negative Numbers

A long time ago, back in February of 2008, when Mathmonsoon had just begun, I discovered an identity that expresses the logarithm with any base (other than -1, 1 or 0) of any number (excluding 0). I was excited because I had always been curious about how my TI-84 calculator and Mathematica are able to spit out logs of negative numbers: for example, they can calculate the log of -4 as some number 𝑎+𝑏𝑖, which turns out to be approximately 1.38629+ 3.14159𝑖. (Note: In the calculations below, if a log has an unspecified base, assume I am talking about the natural log: base 𝑒.) , where 𝑙𝑜𝑔𝑧 (𝑦) is Here’s the formula: 𝑙𝑜𝑔𝑎 (−𝑛) = 𝑙𝑜𝑔𝑎 (𝑛) + 𝑙𝑜𝑔(𝑐𝑖𝑠(𝜋𝑙𝑜𝑔(𝑎))) 𝑙𝑜𝑔(𝑛) 𝑖𝑧 the log base 𝑧 of 𝑦 and 𝑐𝑖𝑠(𝑧) = 𝑒 = 𝑐𝑜𝑠(𝑧)+𝑖𝑠𝑖𝑛(𝑧). Not only is it unusually long and strange-looking, but it doesn’t really seem correct. We can reduce it quite a bit: first, note that 𝑐𝑖𝑠(𝜋𝑙𝑜𝑔(𝑎)) = 𝑒𝑖𝜋𝑙𝑜𝑔(𝑎) , so 𝑙𝑜𝑔(𝑐𝑖𝑠(𝜋𝑙𝑜𝑔(𝑎))) = 𝑖𝜋𝑙𝑜𝑔(𝑎). Then we are left with 𝑙𝑜𝑔𝑎 (−𝑛) = 𝑙𝑜𝑔𝑎 (𝑛) + 𝑖𝜋𝑙𝑜𝑔(𝑎) . Moving the 𝑙𝑜𝑔(𝑛) logs base 𝑎 to the same side we have 𝑙𝑜𝑔𝑎 (−1) = need to notice that 𝑙𝑜𝑔𝑎 (−1) = with

𝑖𝜋 𝑙𝑜𝑔(𝑎)

=

𝑖𝜋𝑙𝑜𝑔(𝑎) , 𝑙𝑜𝑔(𝑛)

𝑖𝜋 𝑙𝑜𝑔(𝑎)

since 𝑎

𝑖𝜋 𝑙𝑜𝑔(𝑎)

𝑖𝜋𝑙𝑜𝑔(𝑎) . 𝑙𝑜𝑔(𝑛) 𝑖𝜋

=𝑒

To continue, we

= −1, so we are left

or 𝑙𝑜𝑔(𝑎)2 = 𝑙𝑜𝑔(𝑛), which clearly isn’t always true.

There are two things that are really weird about this. First, I have no idea where the formula came from. I’ve been messing around with Euler’s formula in every way possible to recreate it, but have come up with nothing but some other cool derivations. The other weird thing is that there is a very simple way to determine the log of a negative number: say we want to find the value of 𝑙𝑜𝑔(𝑧). Then write 𝑧 = 𝑟𝑒𝑖𝜃 so that 𝑙𝑜𝑔(𝑧) = 𝑙𝑜𝑔(𝑟) + 𝑖𝜃. All you need to do now is find 𝜃, which will simply be the angle between the √ 𝑥 and 𝑦 components of the 𝑧 in the complex √ plane. As 𝜋an example, 𝑖𝜋 𝑙𝑜𝑔( 3 + 𝑖) = 𝑙𝑜𝑔(2) + 6 because in this case, 3 + 𝑖 = 2𝑐𝑖𝑠 6 , which you can check yourself. Since we can now take the log of any complex number base 𝑒, it is easy to take the log base anything else using the change of base . formula, 𝑙𝑜𝑔𝛼 (𝛽) = 𝑙𝑜𝑔(𝛼) 𝑙𝑜𝑔(𝛽) Despite my misguided explorations, I’ll go over a few of the cool results I came across. Start with 𝑒𝑖𝜙 = 𝑐𝑜𝑠(𝜙) + 𝑖𝑠𝑖𝑛(𝜙). Then, inverting both sides 1 gives 𝑒−𝑖𝜙 = (𝑐𝑜𝑠(𝜙) + 𝑖𝑠𝑖𝑛(𝜙))−1 = 𝑐𝑜𝑠(𝜙)+𝑖𝑠𝑖𝑛(𝜙) . Multilpying the fraction by 𝑐𝑜𝑠(𝜙)−𝑖𝑠𝑖𝑛(𝜙) 𝑐𝑜𝑠(𝜙)−𝑖𝑠𝑖𝑛(𝜙)

𝑐𝑜𝑠(𝜙)−𝑖𝑠𝑖𝑛(𝜙) gives 𝑒𝜙 = 𝑐𝑜𝑠(𝜙) 2 +𝑠𝑖𝑛(𝜙)2 = 𝑐𝑜𝑠(𝜙) − 𝑖𝑠𝑖𝑛(𝜙). Then, noticing that 𝑖𝜙 𝑐𝑜𝑠(𝜙) − 𝑖𝑠𝑖𝑛(𝜙) = 𝑒 − 2𝑖𝑠𝑖𝑛(𝜙), we have 𝑒−𝑖𝜙 = 𝑒𝑖𝜙 − 2𝑖𝑠𝑖𝑛(𝜙), and solving 𝑖𝜙 −𝑖𝜙 for 𝑠𝑖𝑛(𝜙) gives 𝑠𝑖𝑛(𝜙) = 𝑒 −𝑒 , which is a nice analytic continuation of 2𝑖

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the 𝑠𝑖𝑛 function. I have used this formula in the past, but never derived it for myself. (The 𝑐𝑜𝑠 and 𝑡𝑎𝑛 functions come about from the same method.) Another funky calculation that came out of this was the exact value of 𝑖 , which actually turns out to be a real number! We begin with 𝑖𝑖 = 𝑒𝑖𝑙𝑜𝑔(𝑖) . √ = 𝑖𝜋2 = 𝑙𝑜𝑔( −1) = We know that 𝑒𝑖𝜋 = −1 so that 𝑙𝑜𝑔(−1) = 𝑖𝜋, or 𝑙𝑜𝑔(−1) 2 𝑖𝜋 𝜋 𝑙𝑜𝑔(𝑖), so that 𝑙𝑜𝑔(𝑖) = 𝑖𝜋2 . Then the exact value of 𝑖𝑖 is 𝑒𝑖 2 = 𝑒− 2 . We 𝜋 𝜋 can use this to get the value of things like 𝑖−𝑖 = 𝑒 2 or (𝑖𝑖 )𝑖 = (𝑒− 2 )𝑖 = 𝑖𝜋 𝑒− 2 = 𝑐𝑜𝑠(− 𝜋2 ) + 𝑖𝑠𝑖𝑛(− 𝜋2 ) = −𝑖; but why stop there? What other insane calculations can we come up with?! Well, if (𝑖𝑖 )𝑖 = −𝑖 then ((𝑖𝑖 )𝑖 )𝑖 = (−𝑖)𝑖 = 𝜋 𝜋 𝑒 2 , and (((𝑖𝑖 )𝑖 )𝑖 )𝑖 ) = (𝑒 2 )𝑖 = 𝑐𝑜𝑠( 𝜋2 ) + 𝑖𝑠𝑖𝑛( 𝜋2 ) = 𝑖, and we have reached a periodic point. And that’s pretty sweet: just as 𝑖𝑛 is periodic with period 4, (𝑖𝑘 = 𝑖𝑘+4 ), the power tower of 𝑖’s has period 5. 𝑖

𝜋𝑖

There is another simpler identity that I was able to derive: −𝑛 = 𝑛1+ 𝑙𝑜𝑔(𝑛) . In the real numbers, 𝑛𝑥 = −𝑛 has no solutions for 𝑛 (aside from 𝑛 = 0) with 𝑥 given; we can see this by manipulating to obtain 𝑛𝑥 + 𝑛 = 0, or 𝑛(𝑛𝑥−1 + 1) = 0, and 𝑛𝑥−1 = −1 clearly has no real solutions. However, the above formula gives us the complex power we can raise any number to to obtain its negative. And it is derived very simply: start with Euler’s identity, 𝜋𝑖𝑙𝑜𝑔(𝑛) 𝑙𝑜𝑔𝑛 = −1, power, to obtain 𝑒 𝑒𝜋𝑖 = −1. Then, raise both sides to the 𝑙𝑜𝑔(𝑛) 𝑙𝑜𝑔(𝑛) 𝜋𝑖

𝜋𝑖

or (𝑒𝑙𝑜𝑔(𝑛) ) 𝑙𝑜𝑔(𝑛) = −1. Since 𝑒𝑙𝑜𝑔(𝑛) = 𝑛, we have that 𝑛 𝑙𝑜𝑔(𝑛) = −1, and 𝜋𝑖 multiplying both sides by 𝑛 we are left with −𝑛 = 𝑛 𝑙𝑜𝑔(𝑛) +1 . For example, 𝜋𝑖 2 𝑙𝑜𝑔(2) = −2, a strange result. As usual, I can’t imagine any of these have practical applications; regardless, it’s nice to wrap up an old idea.

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