Sagot :
Answer:
1-i and -1+i
Step-by-step explanation:
We are to find the square roots of [tex]z=0-2i[/tex]. First, convert from Cartesian to polar form:
[tex]r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2[/tex]
[tex]\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}[/tex]
[tex]z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})[/tex]
Next, use the formula [tex]\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr][/tex] where [tex]\displaystyle k=0,1,2,...\:,n-1[/tex] to find the square roots:
When k=1
[tex]\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr][/tex]
[tex]\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr][/tex]
[tex]\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)[/tex]
[tex]\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i[/tex]
When k=0
[tex]\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr][/tex]
[tex]\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)[/tex]
[tex]\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i[/tex]
Thus, the square roots of -2i are 1-i and -1+i