Fundamental Theorem of Algebra

Mathematics Class XI Chapter 2: Complex Numbers and Quadratic Equations Fundamental Theorem of Algebra

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$\to$ “Every Polynomial equation of degree $\geq 1$ has at least on root in $\mathbb{C}$ ”

$\to$ More specifically we have, “Every Polynomial equation of degree $n$ has $n$ roots in $\mathbb{C}$ ”

So, now the equation $ax^2+bx+c=0,~~ a,b,c \in \mathbb{C},~ a \neq 0$ has two roots:

$\alpha_1=\dfrac{-1+\sqrt{b^3-4ac}}{2a}, \quad \alpha_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$

$\to$ In a quadratic equation $ax^2+bx+c=0$ with real coefficients $a,b,c$ complex roots occur in conjugate pairs.

Example 1:

Solve: $x^2+x+1=0$

Solution:

$x^2+x+1=0$ , Here, $b^2-4ac=-3$

$\therefore \alpha_1=\dfrac{-1+\sqrt{-3}}{2}, \quad \alpha_2=\dfrac{-1-\sqrt{-3}}{2}$

$\implies \alpha_1=\dfrac{-1+\sqrt{3}i}{2}, \quad \alpha_2=\dfrac{-1-\sqrt{3}i}{2}$

Example 2:

Solve: $\sqrt{5}x^2+x+\sqrt{5}=0$

Here $b^2-4ac=1^2-4\cdot \sqrt{5}\cdot \sqrt{5}=1-20=-19$

$\therefore \alpha_1=\dfrac{-1+\sqrt{-19}}{2\sqrt{5}}, \quad \alpha_2=\dfrac{-1-\sqrt{-19}}{2 \sqrt{5}}$

$\implies \alpha_1=\dfrac{-1+\sqrt{19}i}{2 \sqrt{5}}, \quad \alpha_2=\dfrac{-1-\sqrt{19}i}{2 \sqrt{5}}$

Example 3:

Solve: $x^2+2=0$

$\implies x^2=-2 \implies x= \pm \sqrt{-2}$

$\implies x=+\sqrt{-2}, ~~ -\sqrt{-2}$

$\therefore x=\sqrt{2}i, ~~ -\sqrt{2}i$

$\therefore x= \pm \sqrt{2}i$

Application

(i) General Solution of the equation:

$x^n=1, n$ is a positive integer

$1=\cos 2 \pi k+i \sin 2 \pi k$

$\thereforex=\cos \dfrac{2 \pi k}{n}+i \sin \dfrac{2\pi k}{n}$ , where $k=0,1,2..., n-1$

So, general solution of the equation $x^n=a$ , where ‘ $a$ ‘ is a complex number, and $a=r(\cos \alpha+i \sin \alpha)$ , then the ‘ $n$ ‘ solutions are

$z= r^{\frac{1}{n}}\left[ \cos \left( \dfrac{\alpha}{n}+\dfrac{2k \pi}{n} \right)+i \sin \left( \dfrac{\alpha}{n}+\dfrac{2k \pi}{n} \right) \right]$

where $k=0,1,2,...n-1$

(ii) Finding square roots of a complex number $P+iq, P, q \in R$

Let $z=P+iq=r (\cos ]alpha+i \sin \alpha)$ , where $r=\sqrt{p^2+q^2}$

$r\cos \alpha=P, ~~ r\sin \alpha=q$

So, we have two roots,

$r^{\frac{1}{2}}\left( \sqrt{\dfrac{1+\cos \alpha}{2}} + i \sqrt{\dfrac{1-\cos \alpha}{2}}\right),$

$-r^{\frac{1}{2}}\left(\sqrt{\dfrac{1+\cos \alpha}{2}}+i \sqrt{\dfrac{1-\cos \alpha}{2}}\right)$

$\implies \left( \sqrt{\dfrac{r+r\cos \alpha}{2}}+i \sqrt{\dfrac{r-r\cos \alpha}{2}} \right),$

$-\left( \sqrt{\dfrac{r+r+\cos \alpha}{2}}+i \sqrt{\dfrac{r-r\cos \alpha}{2}} \right)$

Hence $\left( \sqrt{\dfrac{r+P}{2}}+i \sqrt{\dfrac{r-P}{2}} \right)$ and

$-\left( \sqrt{\dfrac{r+P}{2}} +i \sqrt{\dfrac{r-P}{2}}\right)$

Example:

Obtain the square roots of $3+4i$

Solution:

Let $x, y \in R$ , such that $x+iy=\sqrt{3+4i}$ , then $x^2-y^2+i2xy=3+4i$

$\implies x^2-y^2=3, ~~ 2xy=4$

Now, $(x^2+y^2)=(x^2-y^2)^2+4x^2y^2=3^2+4^2=25$

Since $x^2+y^2$ is non-negative, we have $x^2+y^2=\sqrt{25}=5$

$\therefore\begin{Bmatrix} x^2=4,&\text {i.e.,} &x= \pm 2 \\ y^2=1 & \text {i.e.,} & y=\pm 1 \end{Bmatrix}$

Hence the square roots of $3+4i$ are $2+i$ and $-2-i$