COMPLEX NUMBER - 1 (Algebra of Complex Number)
Our motive for the introduction of complex numbers is to make every algebraic equation solvable. Let us consider the equation $z^{2}+4=0$. This equation has no solution in the set of real numbers. There is no real no. $\mathrm{x}$ whose square is -4 . In order to remedy this situation, a new kind of numbers were introduced and were given the name complex nos. Eular was the first to introduce the symbol $\mathrm{i}$ for $\sqrt{-1}$ with the property $\mathrm{i}^{2}=-1$. $\mathrm{i}$ is also called the symbol as the imaginary unit. $\mathrm{i}$ is called an imaginary number.
Powers of $i$
Complex numbers
An expression of the form $\mathrm{x}+\mathrm{iy}$, where $\mathrm{x} \& \mathrm{y}$ are real numbers and $\mathrm{i}$ is a symbol, is called a complex number and usually denoted by $z$.
Addition of complex number
Let $z _{1}=a _{1}+i b _{1} \& z _{2}=a _{2}+i b _{2}$ be two complex number Then their sum $z _{1}+z _{2}$ is defined as the complex number $\left(a _{1}+a _{2}\right)+i\left(b _{1}+b _{2}\right)$
Multiplication of Complex Number
Let $z _{1}=a _{1}+i b _{1} \& z _{2}=a _{2}+i b _{2}$ be two complex number. Then the multiplication of $z _{1}$ with $z _{2}$ is denoted by $z _{1} z _{2}$ and is defined as $\left(a _{1} a _{2}-b _{1} b _{2}\right)+i\left(a _{1} a _{2}+b _{1} b _{2}\right)$.
Division of Complex Number
The division of a complex number $z _{1}$ by a non-zero complex $z _{2}$ is defined as the multiplicaiton of $z _{1}$ by the multiplication inverse of $z _{2}$ and is denoted by $\frac{z _{1}}{z _{2}}$
$\frac{z _{1}}{z _{2}}=\frac{\left(a _{1}+i b _{1}\right)}{\left(a _{2}+i b _{2}\right)}=\frac{\left(a _{1}+i b _{1}\right)}{\left(a _{2}+i b _{2}\right)} \frac{\left(a _{2}-i b _{2}\right)}{\left(a _{2}-i b _{2}\right)}$
$=\frac{\left(a _{1} a _{2}+b _{1} b _{2}\right)}{a _{2}^{2}+b _{2}^{2}}+\frac{i\left(a _{2} a _{1}-b _{1} b _{2}\right)}{a _{2}^{2}+b _{2}^{2}}$
Conjugate of $\mathrm{Z}$
Let $z=a+i b$ be a complex number. Then the conjugate of $z$ is denoted by $z$ and is equal to $a-i b$ thus $\mathrm{z}=\mathrm{a}+\mathrm{ib}$
$\Rightarrow \quad \overline{\mathrm{z}}=\mathrm{a}-\mathrm{ib}$
e.g. if $z=3+4 i$ then $\bar{z}=3-4 i$
Modulus of Z
The modulus of a complex number $x+$ iy is denoted by $|x+i y|$ and is denoted as $|x+i y|=$ $\sqrt{x^{2}+y^{2}}=$ non-negative square root of $x^{2}+y^{2}$ e.g. $z=3-4 i$, then $|z|=\sqrt{(3)^{2}+(-4)^{2}}=5$.
Equality of complex number
Two complex nos $\mathrm{z} _{1}=\mathrm{x} _{1}+\mathrm{i} \mathrm{y} _{1} \& \mathrm{z} _{2}=\mathrm{x} _{2}+\mathrm{iy} _{2}$ are said to be equal if and only if $\mathrm{x} _{1}=\mathrm{x} _{2} \& \mathrm{y} _{1}=\mathrm{y} _{2}$ i.e. $Z _{1}=Z _{2} \Rightarrow \operatorname{Re}\left(z _{1}\right)=\operatorname{Re}\left(z _{2}\right) \& \operatorname{Im}\left(z _{1}\right)=\operatorname{Im}\left(z _{2}\right)$.
Examples
1. Evaluate $\left[i^{-53}+\left\{i^{n}+i^{n+1}+i^{n+2}+i^{n+3}+4\right\}^{\frac{1}{2}} i\right]$
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Solution: $ \begin{aligned} & \text { G.E }=i^{-(4 \times 13+1)}+\left[\mathrm{i}^{\mathrm{n}}\left(1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}\right)+4\right]^{\frac{1}{2}} \cdot \mathrm{i} \\ & =\left[i^{-1}+\left\{i^{n}+(1+i+(-1)+(-i))+4\right\}^{\frac{1}{2}} \cdot i\right] \\ & =\left[i^{-1}+\left\{i^{n}(0)+4\right\}^{\frac{1}{2}} \cdot i\right] \\ & =\left[i^{-1}+(4)^{\frac{1}{2}} \cdot i\right] \\ & \Rightarrow\left[i^{-1}+2 i\right] \\ & \Rightarrow\left[\frac{1}{i}+2 i\right] \Rightarrow \frac{i}{i^{2}}+2 i \Rightarrow-i+2 i=i \end{aligned} $2. Show that the polynomial $x^{4 p}+x^{4 q+1}+x^{4 r+2}+x^{4 s+3}$ is divisible by $\mathrm{x}^{3}+\mathrm{x}^{2}+\mathrm{x}+1$, where $\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s} \in \mathrm{N}$.
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Solution: Let $\mathrm{f}(\mathrm{x})={ } _{x} 4 p+x^{4 q+1}+x^{4 r+2}+x^{4 s+3}$
Now $\mathrm{x}^{3}+\mathrm{x}^{2}+\mathrm{x}+1=\left(\mathrm{x}^{2}+1\right)(\mathrm{x}+1)=(\mathrm{x}+\mathrm{i})(\mathrm{x}-\mathrm{i})(\mathrm{x}+1)$
$\mathrm{f}(-\mathrm{i})=(-i)^{4 p}+(-i)^{(4 q+1)}+(-i)^{(4 r+2)}+(-i)^{(4 s+3)}=+1+(-i)(+1)+(-\mathrm{i})^{2}+(-\mathrm{i})^{3}=1-i-1+\mathrm{i}=0$ $\mathrm{f}(-1)=0$
Thus by division theorem $\mathrm{x}^3+\mathrm{x}^2+\mathrm{x}+1$ is factor of $x^{4 p}+x^{4 q+1}+x^{4 r+2}+x^{4 s+3}$
3. Express $\frac{1}{1-\cos \theta+2 i \sin \theta}$ in the form $\mathrm{x}+\mathrm{iy}$
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Solution: $ \begin{aligned} & \frac{1}{1-\cos \theta+2 i \sin \theta}=\frac{1}{2 \sin ^{2} \frac{\theta}{2}+4 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} i \\ & =\frac{\left(\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2}\left[\sin \frac{\theta}{2}+2 i \cos \frac{\theta}{2}\right]\left[\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}\right]}=\frac{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}\left[\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}\right]} \\ & =\frac{\left(\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}\right)}{\sin \frac{\theta}{2}\left[2 \sin ^{2} \frac{\theta}{2}+8 \cos ^{2} \frac{\theta}{2}\right]} \\ & =\frac{\left(\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}\right)}{\sin \frac{\theta}{2}[1-\cos \theta+4+4 \cos \theta]} \end{aligned} $
$ =\frac{\left(\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}\right)}{\sin \frac{\theta}{2}(5+3 \cos \theta)} $
$=\frac{1}{5+3 \cos \theta}+\frac{-2 \cot \frac{\theta}{2} i}{5+3 \cos \theta}$
4. Find the multiplicative inverse of complex no $3+2 \mathrm{i}$
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Solution:
Let $\mathrm{z}=3+2 \mathrm{i}$
$\frac{1}{z}=\frac{1}{3+2 i}$
$=\frac{3-2 i}{(3+2 i)(3-2 i)}=\frac{3-2 i}{9+4}=\frac{3}{13}-\frac{2}{13} i$
Practice questions
1. If $a<0, b>0$, then $\sqrt{a} \cdot \sqrt{b}$ is equal to
(a). $-\sqrt{|a| b}$
(b). $-\sqrt{|a| \cdot b i}$
(c). $\sqrt{|\mathrm{a}| \mathrm{b}}$
(d). none of these
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Answer: (b)2. The value of the sum $\sum _{n=1}^{13}\left(i^{n}+i^{n+1}\right)$, where $\mathrm{i}=\sqrt{-1}$, is
(a). $i$
(b). $i-1$
(c). $-\mathrm{i}$
(d). $0$
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Answer: (b)3. The smallest positive integral value of $\mathrm{n}$ of which $\left(\frac{1-i}{1+i}\right)^{n}$ is purely imaginary with positive imaginary part, is
(a). $1$
(b). $3$
(c). $5$
(d). none of these
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Answer: (b)4. If $\mathrm{n}$ is an odd integer, $\mathrm{i}=\sqrt{-1}$, then $(1+\mathrm{i})^{6 \mathrm{n}}+(1-\mathrm{i})^{6 \mathrm{n}}$ is equal to
(a). $0$
(b). $2$
(c). $-2$
(d). none of these
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Answer: (a)5. If $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number and $0<\theta<2 \pi$, then $\theta=$
(a). $\pi$
(b). $\frac{\pi}{2}$
(c). $\frac{\pi}{2}$
(d). $\frac{\pi}{6}$
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Answer: (a)6. If $\mathrm{b}+\mathrm{c}=(1+\mathrm{a}) \mathrm{z}$ and $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1$, then $\frac{1+i z}{1-i z}$
(a). $\frac{a-i b}{1-c}$
(b). $\frac{a-i b}{1+c}$
(c). $\frac{a+i b}{1-c}$
(d). $\frac{a+i b}{1+c}$
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Answer: (d)7. If $(x+i y)^{\frac{1}{3}}=a+i b$ then $\frac{x}{a}+\frac{y}{b}=$
(a). $0$
(b). $1$
(c). $-1$
(d). none of these
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Answer: (d)8. If $(a+i b)^{5}=\alpha+i \beta$, then $(b+a i)^{5}$ is equal to
(a). $\beta+i \alpha$
(b). $\alpha-i \beta$
(c). $\beta-i \alpha$
(d). $-\alpha-i \beta$
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Answer: (a)9. The set of values of $a \in R$ for which $x^{2}+i(a-1) x+5=0$ will have a pair of conjugate complex roots is
(a). $\mathrm{R}$
(b). $\{1\}$
(c). $\left\{a:\left|\cdot a^{2}-2 a+21\right|>0\right\}$
(d). none of these
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Answer: (b)10. The relation between the real numbers $a$ and $b$, which satisfy the equation $\frac{1-i x}{1+i x}=a-i b$, for some real value of $x$, is
(a). $(a-b)(a+b)=1$
(b). $\frac{a-b}{a+b}$
(c). $a^{2}+b^{2}=1$
(d). none of these
