Roots of complex numbers
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HallsofIvy
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First subtract 6+ 6i from both sides: \(\displaystyle z^4= 1+ 3i\).
In "polar form" \(\displaystyle r= \sqrt{1+ 9}= \sqrt{10}\) and \(\displaystyle \theta= arctan(3)= 1.25\) radians (rounded to three significant figures). 1+ 3i can be written as \(\displaystyle \sqrt{10}e^{iarctan(3)\). Using DeMoivre's theorem we take the fourth root of \(\displaystyle \sqrt{10}\), that is \(\displaystyle \sqrt[8]{10}\) which is 1.25 (to the same three significant figures). And we divide the angle 1.25 radians, by 4: 0.322. The first fourth root is \(\displaystyle 1.25e^{0.322i}= 1.25(cos(0.322)+ isin(0.322))\).
But adding \(\displaystyle 2\pi\) to an angle (in radians) just goes once around the whole circle and returns us to the same position. Just as we can write 1+ 3i as \(\displaystyle \sqrt{10}e^{iarctan(3)}\) we can also write it as \(\displaystyle \sqrt{10}e^{I(arctan(3)+ 2\pi)}\) or \(\displaystyle \sqrt{10}e^{i(arctan(3)+ 4\pi)}\) or \(\displaystyle \sqrt{10}e^{i(arctan(3)+ 6\pi)}\)- and we can continue that as far as we wish.
But when we divide the angle by 4, we break that. \(\displaystyle \arctan(3)\) is the same point on the unit circle as \(\displaystyle \arctan(3)+ 2\pi\), \(\displaystyle \arctan(3)+ 4\pi\), \(\displaystyle \arctan(3)+ 6pi\), etc. but \(\displaystyle \arctan(3)/4\), \(\displaystyle (\arctan(3)+ 2\pi)/4= \arctan/4+ \pi/2\), \(\displaystyle (\arctan(3)+ 4\pi)/4= \arctan(3)/4+ \pi\), \(\displaystyle (\arctan(3)+ 6\pi)/4= \arctan(x)/4+ 3\pi/2\) are NOT! It is not until we get to \(\displaystyle \arctan(x)+ 8\pi\) that we get \(\displaystyle (\arctan(x)+ 8\pi)/4= \arctan(x)+ 2\pi\) and are back to the same place on the unit circle.
In "polar form" \(\displaystyle r= \sqrt{1+ 9}= \sqrt{10}\) and \(\displaystyle \theta= arctan(3)= 1.25\) radians (rounded to three significant figures). 1+ 3i can be written as \(\displaystyle \sqrt{10}e^{iarctan(3)\). Using DeMoivre's theorem we take the fourth root of \(\displaystyle \sqrt{10}\), that is \(\displaystyle \sqrt[8]{10}\) which is 1.25 (to the same three significant figures). And we divide the angle 1.25 radians, by 4: 0.322. The first fourth root is \(\displaystyle 1.25e^{0.322i}= 1.25(cos(0.322)+ isin(0.322))\).
But adding \(\displaystyle 2\pi\) to an angle (in radians) just goes once around the whole circle and returns us to the same position. Just as we can write 1+ 3i as \(\displaystyle \sqrt{10}e^{iarctan(3)}\) we can also write it as \(\displaystyle \sqrt{10}e^{I(arctan(3)+ 2\pi)}\) or \(\displaystyle \sqrt{10}e^{i(arctan(3)+ 4\pi)}\) or \(\displaystyle \sqrt{10}e^{i(arctan(3)+ 6\pi)}\)- and we can continue that as far as we wish.
But when we divide the angle by 4, we break that. \(\displaystyle \arctan(3)\) is the same point on the unit circle as \(\displaystyle \arctan(3)+ 2\pi\), \(\displaystyle \arctan(3)+ 4\pi\), \(\displaystyle \arctan(3)+ 6pi\), etc. but \(\displaystyle \arctan(3)/4\), \(\displaystyle (\arctan(3)+ 2\pi)/4= \arctan/4+ \pi/2\), \(\displaystyle (\arctan(3)+ 4\pi)/4= \arctan(3)/4+ \pi\), \(\displaystyle (\arctan(3)+ 6\pi)/4= \arctan(x)/4+ 3\pi/2\) are NOT! It is not until we get to \(\displaystyle \arctan(x)+ 8\pi\) that we get \(\displaystyle (\arctan(x)+ 8\pi)/4= \arctan(x)+ 2\pi\) and are back to the same place on the unit circle.
