# Complex Numbers - 5

ILLUSTRATION - 9.

Find the ecentricity of elipse whose equation is |z - 4| + |z - ^{12}⁄_{5}| = 10 Ans:- Comparing with givem from we get

z_{1} = ae = 4 1^{st} focus

z_{2} = be = ^{12}⁄_{5} 2^{nd} focus

& 2a = 10

=> a = 5

=> ae = 4

5e = 4

=> e = ^{4}⁄_{5} Ans.

1. Iota (i) is neither 0 nor greater than 0, nor less.

2. Amp(z) - Amp(-2) depends as amp(z) positive or negative

3. Segment joining points z_{1} & z_{2}is divided by point z in the ratio m_{1} : m_{2} then

where m_{1} m _{2} R

4. Three point z_{1,} z _{2}, z_{3} collinear if 0

5. is the equation of circle with diameter AB where A ( z_{1}) & B

6. & is aline if k =

7. For evaluation (a + b)^{c + id} we write a + ib^{c + id} = e ^{logea +} = e^{c + id} log_{e}^{a + ib}

8. If avg are concyclic.

__ILLUSTRATION - 10.__

Show that the centre of the circle whose two diametrically opposite point are z_{1} & z_{2} is

Ans. We know taht of a point z divides z_{1} & in ratio m : n then z =

Here m = n = 1 Therefore

Q. 1. Find the value of for while a real volue of x will satisfy the equation

__Ans__:-

So,

Equating real & Imaginary parts of equation

Removing x from above two equation we get ,which is an identity & true for every .

Q.2. Rind the square root of 15 + 8i.

__Ans__ Let the square root of 15 + 8i be a + 1 Then = a + bi squaring both

we get 15 + 8i = a_{2} - b_{2} & 8 = 2ab Solving above two equation a = 4, b = Hence square root of 15 + 8i is 4 + i

Q.3. …………

__Ans__ If z =

then

Therefore

modules of e^{i} = 1

Argument of e^{i} =

Q. 4. Find the value of

Ans:- We know that

Henc

Therefore

Q.5. If z_{1} & z_{2} are two complex number, then find the value of k in the equation |z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + ki ?

Ans:- We know that,

|z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + 2 Re

Comparing with the given form we get,

|z_{1}|^{2} + |z_{2}|^{2} + ki = |z_{1}|^{2} + |z_{2}|^{2} + 2 Re

=> Ki = Real

So k should be purely Imaginary Number.

Q. 6. If iz^{3} + z^{2} - z + i = 0 . Find modulus of z.

Ans:- iz^{3} + z^{2} - z + i = 0

iz^{3} + z^{2} + i^{2}z + i = 0

z^{2}(iz + 1) + i(iz + 1) = 0

(z^{2} + i)(iz + 1) = 0

Solution are z =i & z^{2} = - i

Considering z = i Taking modulus on both sides

|z| = |i| = 1

Now consider z^{2} = -i, Take modulus of both sides

|z^{2}| = |-i| = 1

|z^{2}| = |z|^{2} = 1

=> |z| = 1 but |z| always

Hence |z| = 1

Q.7. ……………

number z satisfying |z - i| = 4

Ans:- We know taht |z_{1} -|z_{2}| |z_{1}-z_{2}| |z_{1}-z_{2}

Here z_{1} & z_{2} = i

So |z| - |i| |z - i| |z| + |i|

|z| - 1 4 |z| + 1

|z| - 1 => |z| 5

|z| + 1 => |z| 3

Hence 3 |z| 5

Q.8. Find the locus of z Satisfying lthe equation Arg

Ans:- Let z = x + iy yhen Arg = Arg(z - 1) - Arg (z + 1)

= Arg((x - 1) + iy) - Arg ((+ 1) + iy)

Locus is a circle with centre(0, 1)& radius.

Q.9. If be roots of equation x^{5} - 1 Then find the value of

Now we know lthat

In this equation replacing x by - x we get

Now in this equation replacing x by we get

Hence the value is

Q. 10. Find the condition so that complex numbers z_{1} & z_{2} & origin form an isosceles triangle with vwetical angle .

Ans:- In this case |z_{1}| = |z_{2}|

A = |z_{1}|

B = |z_{2}|

Applying Rotation thesrem to this trianglem takin origin at O we get

Squaring both side & rearecanging we get

z_{2}^{2}+ z_{1}^{2} +

Q.11. Show that the area of the triangle of the argand digram formed by iz, -z & -z + iz is |z|^{2} ?

Ans:- It is clear from the vector theory that the Right angled triangle can be made through these three comples number. Area of Right angled triangle = X base X height

base = |iz| = |z|

Height = |i^{2}z| = |-z| = |z|

Area = X |z| X |z| = |z|^{2}

__Dumb Question__:- Why this D is a right angled ?

Ans:- Take any complex no. as z. If you multiply it with i we get iz = ze^{iz/2} If Hence iz is the complex no. having same modulus as z but amplitude is 90 more than z.

Q.1. If w is the nth root of unity & z_{1} & z_{2} are any two complex number prove that

( n N)

Ans:- If 1, w, w^{2}, …….w^{n - 1}are n nth roots of unity then

Now consider

So,

Hence proved

__Dumb Question__:- If then how .

Ans:- If Take conjugate of both side

Q. 3. Prove that xy(x + y)(x^{2} + y^{2} xy) is a factor of g(x, y) = (x + y)^{n} - x^{n} - y^{n}

Ans:- (xy)(x +y)(x^{2} + y^{2} + xy) can be writen as

f(x, y) = (xy)(x + y)(x - wy)(x - w^{2}y)

If we can prove taht whenever f(x,y) = 0, g(x,y) = (x + y)^{n} - x^{n} - y ^{n} is also 0, then f(x,y) is always a factor of (x + y)^{n} - x^{n} - y^{n}

1. Put x = 0 which is afactor of f(x,y) in g(x,y)

we get g(0, y) = (0 + y)^{n} - 0^{n} - y^{n}

= y^{n} - y ^{n} = 0

i.e. (x - 0) is a factor of g(x,y)

2. Put y = 0 whichis a factor of f(x,y) in g(x,y)

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

= x^{n} - x^{n} = 0

i. e. (y - 0) is also a factor of g(x, y)

3. Put (x + y) = 0 which is a factor of f(x, y) in g(x, y)

g(x, y)/x + y = 0 = 0^{n} - (- y)^{n} - y^{n}

= - (- y)^{n} - y^{n} (becaise n is 0

i.e (x + y) is also a factor of g(x, y)

4. Put x = wy which is afactor of f(x, y) in g(x, y)

g(wy, y) = (wy + y)^{n} - (wy)^{n} - y^{n}

= y^{n} [- w^{2n} - w^{n} - 1]

g(wy, y) = - y^{n}(w^{2n} + w + 1)

n is not a multiple of 3, n can be 3n + 1 or 3n in bot the cases w^{2n} + w^{n} + 1 = 0, Hence g(wy, y) = 0

i.e. (x - wy) is also a factor of g(x, y)

5. Put x = w^{2}y which is a factor of f(x, y) in g(x, y)

g(w^{2}y, y) = (w^{2}y + y)^{n} - (w^{2}y)^{n} - y^{n}

= y^{n}(- w^{2} - w^{2n} - 1)

= - y^{n}(1 + w^{n} + w^{2n})

g(w^{2}y, y) = 0

i.e. (x - w^{2}y ) is also a factor of g(x, y) combing all the above factor which are common to both we can say xy(x + y)(x^{2} + xy + is a factor of g(x, y) = (x + y)^{n} - x^{n} - y^{n}

Q.3. Interpret the equation geometrically on the Argand plane :

Ans:-

S0,

Put |z - 1| = x

So x > 8 & x < ^{4}⁄_{3}

Hence |z - 1| > 8 represents th exterior of a circle with centre (1, 0) & radius 8 & |z - 1| < ^{4}⁄_{3} represents the interior of a circle with centre (1, 0) & radius ^{4}⁄_{3} :

__Dumb Question__:- Why K < ^{1}⁄_{2} when log_{1⁄2}K > 1

Ans : log_{1⁄2}K > 1 means. log_{1⁄2}K > log_{1⁄2}K > ^{1}⁄_{2} And we know that when the base is less than 1 the

inequality cahnges. Hence K < ^{1}⁄_{2}

Q.4. Diwaing f(z) by z - 2i we ge remainder i diwding by z + 2i we get the remainder 2i. Find the remainder upon division of z^{2} + 4 ?

Ans:- gives remainder as i means

f(z) = K(z - 2i) + i

i.e. f(2i) = i

& f(-2i) = 2i