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Zigmanuir [339]
3 years ago
9

Express the following complex number in trigonometric form: 3 - 3i and find the 4th roots.

Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Answer:

z=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right)

z_1=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{7\pi}{16}+i\sin\dfrac{7\pi}{16}\right).

z_2=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{15\pi}{16}+i\sin\dfrac{15\pi}{16}\right).

z_3=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{23\pi}{16}+i\sin\dfrac{23\pi}{16}\right).

z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).

Step-by-step explanation:

The complex number z=3-3i has the real part Re\ z=3 and the imaginary part Im\ z=-3.

Hence,

|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}.

From the last two equalities, \varphi =\dfrac{7\pi}{4} and the trigonometric form is

z=|z|(\cos\varphi+i\sin\varphi)=3\sqrt{2}\left(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4}\right).

The square roots can be calculated using the formula:

\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\}.

At k=0:

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).

At k=1:

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).

At k=2:

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).

At k=3:

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).

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