Some Solved Examples-I: Complex Number

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Ques 1: Find the number of solutions of the equations, $\arg .\left( {\frac{{z - 2 - i}}{{z - 4 - 3i}}} \right) = \frac{\pi }{4}$ and |z - 3 + i| = 3 ?

Sol:
1. Equation $\arg .\left( {\frac{{z - 2 - i}}{{z - 4 - 3i}}} \right) = \frac{\pi }{4}$ is an equation of arc of the circle. The angle should be measured from denominator z - 4 - 3i to numerator z - 2 - i.

End points of the arc (chord) of the circle are (4, 3) and (2, 1)

distance between (4,3) and (2,1) = 2 root 2

Angle at centre = pi/2

so, radius = 2

Centre: (2,2)

2. Equation |z - 3 + i| = 3 is an equation of circle.

Centre (3, -1) and radius = 3

Draw the above graph of circle,

find the position of (2,1) in respect of the circle,

|2 + i - 3 + i| - 3 = root 5 - 3 < 0

therefore point (2,1) lies inside the circle |z - 3 + i| = 3

Therefore, from the figure it is clear that both the graphs cut each other only on one point.
Therefore, there is only one common point on both the curve. So, there is only one solution exist for the given equations

Ques No. 2: If $\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0$ then prove that
$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos (\alpha + \beta + \gamma )$
and $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = \sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0$

Solution:
Define, $x = \cos \alpha + i\sin \alpha = {e^{i\alpha }}$
$y = \cos \beta + i\sin \beta = {e^{i\beta }}$
$z = \cos \gamma + i\sin \gamma = {e^{i\gamma }}$

Now, $x + y + z = \cos \alpha + \cos \beta + \cos \gamma + i(\sin \alpha + \sin \beta + \sin \gamma )$

or,

therefore, ${x^3} + {y^3} + {z^3} = 3xyz$

${(\cos \alpha + i\sin \alpha )^3} + {(\cos \beta + i\sin \beta )^3} + {(\cos \gamma + i\sin \gamma )^3}$
$= 3(\cos \alpha + i\sin \alpha )(\cos \beta + i\sin \beta )(\cos \gamma + i\sin \gamma )$

by De Moiver's theorem,

$\cos 3\alpha + \cos 3\beta + \cos 3\gamma + i(\sin 3\alpha + \sin 3\beta + \sin 3\gamma ) = 3[\cos (\alpha + \beta + \gamma ) + i\sin (\alpha + \beta + \gamma )]$

Comparing real and imaginary values from both sides we get,

$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos (\alpha + \beta + \gamma )$ (Hence Proved)

Next, ${(x + y + z)^2} = {x^2} + {y^2} + {z^2} + 2(xy + yz + zx)$

$0 = {x^2} + {y^2} + {z^2} + 2(xy + yz + zx)$

or, ${x^2} + {y^2} + {z^2} + 2xyz(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 0$

[ since ${\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = {x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}}$

${ = {{(\cos \alpha + i\sin \alpha )}^{ - 1}} + {{(\cos \beta + i\sin \beta )}^{ - 1}} + {{(\cos \gamma + i\sin \gamma )}^{ - 1}}}$

${ = \cos \alpha + \cos \beta + \cos \gamma - i(\sin \alpha + \sin \beta + \sin \gamma )}$

= 0 ]

therefore, ${{x^2} + {y^2} + {z^2} + 2xyz.0 = 0}$

or, ${{x^2} + {y^2} + {z^2} = 0}$

${{{(\cos \alpha + i\sin \alpha )}^2} + {{(\cos \beta + i\sin \beta )}^2} + {{(\cos \gamma + i\sin \gamma )}^2} = 0}$

or, ${\cos 2\alpha + \cos 2\beta + \cos 2\gamma + i(\sin 2\alpha + \sin 2\beta + \sin 2\gamma ) = 0}$ + 0i

Compare both sides,

$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = \sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0$ (hence proved)

Ques 3: If n is an odd integer but not a multiple of 3 then prove that $xy(x + y)({x^2} + {y^2} + xy)$ is a factor of ${(x + y)^n} - {x^n} - {y^n}$ ?

Sol:
We can write,

$xy(x + y)({x^2} + {y^2} + xy)$ = $xy(x + y)(x - \omega y)(x - {\omega ^2}y)$

So, If $xy(x + y)({x^2} + {y^2} + xy)$ is to be a factor of ${(x + y)^n} - {x^n} - {y^n}$ then x, y , (x + y), $(x - \omega y)$ , and $(x - {\omega ^2}y)$ are also the factors of ${(x + y)^n} - {x^n} - {y^n}$ ( where $\omega$ and ${\omega ^2}$ are the roots of unity)

Let $f(x,y) = {(x + y)^n} - {x^n} - {y^n}$

So, for x to be factor, put x = 0

$f(0,y) = {(0 + y)^n} - {0^n} - {y^n} = 0$

Factor means you can write the polynomial in multiplication of some variables and if you put zero in place of any factor or multiplication. the value of polynomial will get zero Therefore x is a factor.

Put y = 0

$f(x,0) = {(x + 0)^n} - {x^n} - {0^n} = 0$

therefore, y is a factor of above polynomial,

Put $x - \omega y = 0$ => $x = \omega y$

$f(\omega y,y) = {(\omega y + y)^n} - {(\omega y)^n} - {y^n}$

$= {(\omega + 1)^n}{y^n} - {\omega ^n}{y^n} - {y^n}$ ( since $1 + \omega + {\omega ^2} = 0$ $\Rightarrow 1 + \omega = - {\omega ^2}$ )

$= [{( - {\omega ^2})^n} - {\omega ^n} - 1]{y^n}$

$= [ - {\omega ^2}^n - {\omega ^n} - 1]{y^n}$ ( since n is odd integer and not multiple of 3 i.e. ${\omega ^n} \ne 1$ )

$= - [{\omega ^2}^n + {\omega ^n} + 1]{y^n}$ = 0 ( since $\omega$ and ${\omega ^2}$ are the roots of unity
$\Rightarrow 1 + {\omega ^n} + {\omega ^{2n}} = 0$ )

therefore, $(x - \omega y)$ is a factor of f(x,y)

Similarly, put $x - {\omega ^2}y = 0 \Rightarrow x = {\omega ^2}y$

$f({\omega ^2}y,y) = {({\omega ^2}y + y)^n} - {({\omega ^2}y)^n} - {y^n}$

= $[{({\omega ^2} + 1)^n} - {\omega ^{2n}} - 1]{y^n}$ ( since $1 + {\omega ^2} = - \omega$ )

$= - [{\omega ^2}^n + {\omega ^n} + 1]{y^n}$ = 0

Therefore, $xy(x + y)({x^2} + {y^2} + xy)$ = $xy(x + y)(x - \omega y)(x - {\omega ^2}y)$ is a factor of ${(x + y)^n} - {x^n} - {y^n}$

Ques 4: Show that the polynomial $f(x) = {x^{4p}} + {x^{4q + 1}} + {x^{4r + 2}} + {x^{4s + 3}}$ is divisible by ${x^3} + {x^2} + x + 1$ where p, q, r and s are positive integers ?

Sol: Let $g(x) = {x^3} + {x^2} + x + 1$ $= x({x^2} + 1) + ({x^2} + 1)$
$= ({x^2} + 1)(x + 1)$


Now for x - i to be a factor, put x = i

Therefore,

$f(i) = {i^{4p}} + {i^{4q + 1}} + {i^{4r + 2}} + {i^{4s + 3}} = 1 + {i^{4q}}i + {i^{4r}}{i^2} + {i^{4r}}{i^3}$

   = 0 ( if power of 'i' is a multiple of 4 it will give the value equal to 1)

Similarly, f(-i) = 0 , ( x+ i) is a factor

and f(1) = 0, (x - 1) is a factor

Therefore, $f(x) = {x^{4p}} + {x^{4q + 1}} + {x^{4r + 2}} + {x^{4s + 3}}$ is divisible by $({x^2} + 1)(x + 1) = (x + i)(x - i)(x + 1)$

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Some Solved Examples-I: Complex Number

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