Question

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

Answer 1

`z=3√(2)\left(\cos(7\pi)/(4)+i\sin(7\pi)/(4)\right)`

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

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

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

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

Explanation:

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

Hence,

`|z|=√((Re\ z)^2+(Im\ z)^2)=√(3^2+(-3)^2)=√(9+9)=3√(2),\\ \\\cos \varphi=(Re\ z)/(|z|)=(3)/(3√(2))=(√(2))/(2),\\ \\\sin \varphi=(Im\ z)/(|z|)=(-3)/(3√(2))=-(√(2))/(2).`

From the last two equalities,
`\varphi =(7\pi)/(4)` and the trigonometric form is

`z=|z|(\cos\varphi+i\sin\varphi)=3√(2)\left(\cos(7\pi)/(4)+i\sin(7\pi)/(4)\right).`

The square roots can be calculated using the formula:

`\sqrt[4]{z}=\left\{\sqrt[4]z\left(\cos(\varphi+2\pi k)/(4)+i\sin(\varphi+2\pi k)/(4)\right),\text{ where }k=0,1,2,3\right\}.`

At k=0:

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

At k=1:

`z_2=\sqrt[4]{3√(2)}\left(\cos((7\pi)/(4)+2\pi)/(4)+i\sin((7\pi)/(4)+2\pi)/(4)\right)=\sqrt[4]{3√(2)}\left(\cos(15\pi)/(16)+i\sin(15\pi)/(16)\right).`

At k=2:

`z_3=\sqrt[4]{3√(2)}\left(\cos((7\pi)/(4)+4\pi)/(4)+i\sin((7\pi)/(4)+4\pi)/(4)\right)=\sqrt[4]{3√(2)}\left(\cos(23\pi)/(16)+i\sin(23\pi)/(16)\right).`

At k=3:

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