91影视

Given expression ${\left[1-\sqrt{3i}\right]}^6$:

We have to write it in the standard form.

To solve a complex number raised to (n) it is necessary that it be written in polar form, therefore for the given number we have to :

$z^6={\left[1-\sqrt{3i}\right]}^6$

We know that a complex number can be written in standard form as a + bi now to convert this number into polar f贸rm

$z=(r\cos 胃+i\sin 胃)$

$\begin{array}{r}r=\sqrt{x^2+y^2}\\ x=r\cos 鈦�胃\\ y=r\sin 鈦�胃\end{array}$

$\begin{array}{r}r=\sqrt{1^2+(鈭�\sqrt{3}{)}^2}\\ =\sqrt{1+3}\\ =2\end{array}$

$\begin{array}{r}x=r\cos 鈦�胃\\ \cos 鈦�胃=\frac{x}{r}\\ 胃={\cos }^{鈭�1}鈦�\frac{x}{r}\\ 胃={\cos }^{鈭�1}鈦�\frac{1}{2}\\ ={60}^{鈭�}\end{array}$

It can be rewritten as:

$胃={300}^{鈭�}=\frac{5\mathrm{蟺}}{3}$

Now we can write the polar form of the given complex number as:

$z=2\left(\frac{5蟺}{3}+i\sin 鈦�\frac{5蟺}{3}\right)$

To convert complex number raised to the power n,in the standard form, we know that:

$z^n=r^n(\cos 胃+i\sin 胃)$

$\begin{array}{r}{\mathbf{z}}^6={\left[2\left(\cos 鈦�\frac{5蟺}{3}+\mathbf{i}\sin 鈦�\frac{5蟺}{3}\right)\right]}^6\\ =2^6\left[\cos 鈦�\left(6鈰�\frac{5蟺}{3}\right)+\mathbf{i}\sin 鈦�\left(6鈰�\frac{5蟺}{3}\right)\right]\\ =64(\cos 鈦�10蟺+\mathbf{i}\sin 鈦�10蟺)\\ =64(1+\mathbf{i}0)\\ =64\end{array}$