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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
avatar+516 
+4

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
avatar+488 
0

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

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