Answer:
[tex]z=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right)[/tex]
[tex]z_1=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{7\pi}{16}+i\sin\dfrac{7\pi}{16}\right).[/tex]
[tex]z_2=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{15\pi}{16}+i\sin\dfrac{15\pi}{16}\right).[/tex]
[tex]z_3=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{23\pi}{16}+i\sin\dfrac{23\pi}{16}\right).[/tex]
[tex]z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).[/tex]
Step-by-step explanation:
The complex number [tex]z=3-3i[/tex] has the real part [tex]Re\ z=3[/tex] and the imaginary part [tex]Im\ z=-3.[/tex]
Hence,
[tex]|z|=\sqrt{(Re\ z)^2+(Im\ z)^2}=\sqrt{3^2+(-3)^2}=\sqrt{9+9}=3\sqrt{2},\\ \\\cos \varphi=\dfrac{Re\ z}{|z|}=\dfrac{3}{3\sqrt{2}}=\dfrac{\sqrt{2}}{2},\\ \\\sin \varphi=\dfrac{Im\ z}{|z|}=\dfrac{-3}{3\sqrt{2}}=-\dfrac{\sqrt{2}}{2}.[/tex]
From the last two equalities, [tex]\varphi =\dfrac{7\pi}{4}[/tex] and the trigonometric form is
[tex]z=|z|(\cos\varphi+i\sin\varphi)=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right).[/tex]
The square roots can be calculated using the formula:
[tex]\sqrt[4]{z}=\left\{\sqrt[4]{|z|}\left(\cos\dfrac{\varphi+2\pi k}{4}+i\sin\dfrac{\varphi+2\pi k}{4}\right),\text{ where }k=0,1,2,3\right\}.[/tex]
At k=0:
[tex]z_1=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}}{4}+i\sin\dfrac{\frac{7\pi}{4}}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{7\pi}{16}+i\sin\dfrac{7\pi}{16}\right).[/tex]
At k=1:
[tex]z_2=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+2\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+2\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{15\pi}{16}+i\sin\dfrac{15\pi}{16}\right).[/tex]
At k=2:
[tex]z_3=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+4\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+4\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{23\pi}{16}+i\sin\dfrac{23\pi}{16}\right).[/tex]
At k=3:
[tex]z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+6\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+6\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).[/tex]
