Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Fundamental Theorem of Algebra

 \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

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