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# Complex Numbers

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Find all complex numbers z such that z^2 = 2i + 2.

Write your solutions in  form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (Don't include quotes in your answer.)

Feb 16, 2022

#1
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First we have to square root both sides, thus, $$z = \pm\sqrt{2i + 2}$$.

$$\sqrt{2i + 2} = (2i + 2)^{1\over2}$$

Then, opening paranthesis is $$\sqrt{2i} + \sqrt{2} + 2i$$.

That simplifies to $$\sqrt{2} + i\sqrt{2} + \sqrt{2} + 2i$$ and $$-\sqrt{2} - i\sqrt{2} + \sqrt{2} + 2i$$

Thus, the complex numbers of z are $$2i + 2\sqrt{2} +i\sqrt{2}$$ and $$2i - i\sqrt{2}$$.

or maybe im imagining things :)

Feb 16, 2022
#2
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Nice answer proyaop, but squaring either one of those doesn't end up with 2i+2

$$2+2i\\ =2\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ =2\sqrt{2}(\cos(45^\circ)+i\sin(45^\circ))$$

when taking the square root, the argument divides by 2 and the magnitude square roots, so it is:

$$=2^{\frac{3}{4}}(\cos(22.5^\circ)+i\sin(22.5^\circ))$$

To solve for $$\cos(22.5^\circ)$$ and $$\sin(22.5^\circ)$$, we can make use of the half-angle identity:

$$\cos(22.5^\circ)\\= \sqrt{\frac{1+\cos(45^\circ)}{2}}\\= \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2+\sqrt{2}}{4}}\\= \frac{\sqrt{2+\sqrt{2}}}{2}$$

(it's positive because it's in the first quadrant)

Same thing with sin:

$$\sin(22.5^\circ)\\= \sqrt{\frac{1-\cos(45^\circ)}{2}}\\= \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2-\sqrt{2}}{4}}\\= \frac{\sqrt{2-\sqrt{2}}}{2}$$

Thus, the final root is:

$$2^{\frac{3}{4}}(\frac{\sqrt{2+\sqrt{2}}}{2}+\frac{\sqrt{2-\sqrt{2}}}{2}i)\\ =\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}+\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i$$

The simplifying is up to you.

Also, there is one more root, if you use $$\cos(22.5^\circ+180^\circ)$$ and $$\sin(22.5^\circ+180^\circ)i$$, because there are 2 roots of unity. Notice that the cosine and sine flip signs in that case, because they get to the 3rd quadrant. So the other root is $$=-\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}-\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i$$

I seriously doubt that the problem was meant to be this complicated

textot  Feb 16, 2022