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How can I convert the complex number $\mathrm i^\pi$ to trigonometric form?

I usually do these steps:

  1. take $ Z = a + b\mathrm i $ form,
  2. find $ r = \sqrt{a^2 + b^2} $,
  3. $ \cos(\phi) = a / r, \sin(\phi) = b / r $,
  4. find $ \phi $ from the above 2 equations.

For $ \mathrm i^\pi $ I have $ a = 1, b = 1, r = 1 $, $ \cos(\phi) = 1, \sin(\phi) = 1 $. There's no such $ \phi $.

The online Convert Complex Numbers to Polar Form gives the answer $ \phi = 77.2567 $ or just, $ \phi = \dfrac{180 \arg(\mathrm i^\pi)}{\pi}$

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4 Answers 4

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ComplexExpand [$i^\pi$]

$ \cos(\frac{\pi^2}{2}) + i\sin(\frac{\pi^2}{2})$

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  • $\begingroup$ i don't get the logic behind the calculations. where is the pi^2/2 from? $\endgroup$ Nov 6, 2018 at 6:21
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    $\begingroup$ @user3132457 $ \mathrm i = \mathrm e^{\mathrm i \pi/2} $ $\endgroup$ Nov 6, 2018 at 7:26
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Some useful identities:

$i = e^{i\pi/2}$

$(a^b)^c = a^{(bc)}$

$e^{i\phi} = \cos{\phi} + i\sin{\phi}$

Your expression:

$i^\pi = (e^{i\pi/2})^\pi = e^{i\pi^2/2} = 1 (\cos{(\pi^2/2)} + i \sin{(\pi^2/2}))$

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That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:

I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*{x -> (I π^2)/2}*)

Now we can just convert to trig

ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from

{Re[%], Im[%]} // ToPolarCoordinates // N
(*{1., -1.34838}*)

If you like degrees better

{%[[1]], %[[2]]/°}
(*{1., -77.2567}*)
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So, first of all, you made a small mistake : In your method, you have $a = 0$, and not $a=1$ that makes $b=1$, $r=1$, $\cos(\phi)=0$ and $\sin(\phi)=1$, that implies $\phi = \frac{\pi}{2} + 2k\pi$ for any $k \in \mathbb{Z}$.

The problem is all values of $k$ are correct (as $\cos(\alpha) = \cos(\alpha+2\pi)$, whatever the value of $\alpha$, and we have the same for $\sin$), and as such, $i^\pi$ doesn't make any mathematical sense, since $e^{i\frac{\pi^2}{2}} \neq e^{i\frac{5\pi^2}{2}}$, and mathematically speaking, you have no reason to chose one over the other.

Math software usually ignore this by arbitrarily choosing the argument value that is between $-\pi$ and $\pi$ (or sometimes $0$ and $2\pi$, which even there can lead to different results if you took, for instance, "$(-i)^\pi$" )

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  • $\begingroup$ in this case phi is (Pi^2)/2 + 2 Pi k and E^(I*(Pi^2/2)) does indeed equal E^(I*(Pi^2/2 + 2*Pi)). You seem to be adding 2*Pi^2 to phi which is not right. I^Pi makes perfectly good mathematical sense. $\endgroup$
    – Bill Watts
    Nov 6, 2018 at 18:57

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