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Today I read that you can see the number $i$ as the rotation of 90° and therefore i^2 is the rotation of 180° or -1.

I also learned that $1+i$ is 45° but if this would be true I should be able to rotate 4 times with 45° and I should also get to 180° or -1 but $(1+i)^4 = -4$

It seems that $(1+i) * (1+i)$ also scales along the $i$ axes by 2.

I am still new to the imaginary number and I am trying to get an intuition.

Brad
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Maik Klein
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    $(1+i)$ is not purely a direction since it has length $\sqrt{2}$, it is pointing in the direction of the angle $\pi/ 4$. You'll note that -1 and -4 are in the same direction, that much is all you can get, the size can be anything. – Adam Hughes Jul 20 '14 at 21:46
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    What you're missing is a factor of $\frac{1}{\sqrt 2}$. It is true that $\frac{1+i}{\sqrt 2}$ is a rotation by $45$ degrees but what you have is a rotation and a scaling. – Cameron Williams Jul 20 '14 at 21:46
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    Are $!=$ and $\ne$ same? – hola Jul 20 '14 at 21:47
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    Also (joke response for the analysts out there): "Because $i^2-(1+i)^4\ne 0$." – Adam Hughes Jul 20 '14 at 21:50
  • @pushpen.paul Yes. – Arthur Jul 20 '14 at 21:50
  • "I also learned that $1+i$ is $45^\circ$" ... multiplication by a complex number is generally more than just a rotation, it is also a dilation (shrink or expansion) by a factor equal to the "size" (absolute value) of the complex number. – anon Jul 21 '14 at 08:25

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You can write $i+1$ as $\sqrt{2} (\cos 45^{\circ}+i\sin 45^{\circ})$, so by De Moivre's formula $(i+1)^4=(\sqrt{2})^4(\cos 180^{\circ}+i\sin 180^{\circ})$.

Generally, multipling by $\cos \alpha+i \sin \alpha $ is rotation of $\alpha$, multipling by $\beta \in \mathbb{R}$ is scaling by $\beta$.

agha
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Another way of seeing it is that for $\theta = \frac{\pi}{4}, {\frak{Re}} (z) = 1, {\frak{Im}} z = 1$ (i.e. $r = \sqrt{2}$) Euler's expression will be $\sqrt{2} e^{i \frac{\pi}{4}}$, which, after putting it to the power 4 is certainly not equal to $-1$.

Alex
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