Define conjugate of a complex number. Using De Moivre's theorems, find the square root of 2 + 2 3i

Model Answer

1. Definition of Conjugate of a Complex Number

The conjugate of a complex number $z=a+bi$ (where $a,b\in \mathbb{R}$ and $i=\sqrt{-1}$) is defined as: $$\overline{z}=a-bi$$ In other words, the conjugate of a complex number is obtained by changing the sign of its imaginary part.

Example: If $z=3+4i$, then its conjugate is $\overline{z}=3-4i$.

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2. Finding the Square Root of $-2+2\sqrt{3}i$ Using De Moivre’s Theorem

Step 1: Express the complex number in polar form

Let $z=-2+2\sqrt{3}i$.

• Modulus ($r$): $$r=\sqrt{a^2+b^2}=\sqrt{(-2{)}^2+(2\sqrt{3}{)}^2}=\sqrt{4+12}=\sqrt{16}=4$$

• Argument ($\theta$): The complex number lies in the second quadrant (since $a=-2<0$ and $b=2\sqrt{3}>0$). $$\theta =\pi -{\tan }^{-1}\left(\left|\frac{b}{a}\right|\right)=\pi -{\tan }^{-1}\left(\frac{2\sqrt{3}}{2}\right)=\pi -{\tan }^{-1}(\sqrt{3})=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$$

  Thus, the polar form is: $$z=4\left(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right)$$

Step 2: Apply De Moivre’s Theorem for square roots

The square roots of $z$ are given by: $$\sqrt{z}=\sqrt{r}\left[\cos \left(\frac{\theta +2k\pi }{2}\right)+i\sin \left(\frac{\theta +2k\pi }{2}\right)\right],k=0,1$$

• For $k=0$: $${\sqrt{z}}_1=2\left[\cos \left(\frac{2\pi /3}{2}\right)+i\sin \left(\frac{2\pi /3}{2}\right)\right]=2\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)$$ $$=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=1+i\sqrt{3}$$

• For $k=1$: $${\sqrt{z}}_2=2\left[\cos \left(\frac{2\pi /3+2\pi }{2}\right)+i\sin \left(\frac{2\pi /3+2\pi }{2}\right)\right]=2\left(\cos \left(\frac{7\pi }{6}\right)+i\sin \left(\frac{7\pi }{6}\right)\right)$$ $$=2\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)=-\sqrt{3}-i$$

Final Answer: The square roots of $-2+2\sqrt{3}i$ are: $$\boxed{1+i\sqrt{3}}\text{and}\boxed{-\sqrt{3}-i}$$