Question

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

Answer 1

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:

$z_4=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{\frac{7\pi}{4}+6\pi}{4}+i\sin\dfrac{\frac{7\pi}{4}+6\pi}{4}\right)=\sqrt[4]{3\sqrt{2}}\left(\cos\dfrac{31\pi}{16}+i\sin\dfrac{31\pi}{16}\right).$