Complex Numbers - 5 · KnowledgeBin.org

ILLUSTRATION - 9.
Find the ecentricity of elipse whose equation is |z - 4| + |z - 12/5| = 10 Ans:- Comparing with givem from we get
z₁ = ae = 4 1^st focus
z₂ = be = 12/5 2^nd focus
& 2a = 10
=> a = 5
=> ae = 4
5e = 4
=> e = 4/5 Ans.

Iota (i) is neither 0 nor greater than 0, nor less.
Amp(z) - Amp(-2) depends as amp(z) positive or negative
Segment joining points z₁ & z₂is divided by point z in the ratio m₁ : m₂ then
where m₁ m ₂ R
Three point z_1, z ₂, z₃ collinear if 0
is the equation of circle with diameter AB where A ( z₁) & B
& is aline if k =
For evaluation (a + b)^{c + id} we write a + ib^{c + id} = e ^{log_e^{a +}} = e^{c + id} log_e^{a + ib}
If avg are concyclic.

ILLUSTRATION - 10.
Show that the centre of the circle whose two diametrically opposite point are z₁ & z₂ is

Ans. We know taht of a point z divides z₁ & 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 = a₂ - b₂ & 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 eⁱ = 1
Argument of eⁱ =

Q. 4. Find the value of

Ans:- We know that
Henc
Therefore

Q.5. If z₁ & z₂ are two complex number, then find the value of k in the equation |z₁ + z₂|² = |z₁|² + |z₂|² + ki ?

Ans:- We know that,
|z₁ + z₂|² = |z₁|² + |z₂|² + 2 Re
Comparing with the given form we get,
|z₁|² + |z₂|² + ki = |z₁|² + |z₂|² + 2 Re

=> Ki = Real
So k should be purely Imaginary Number.

Q. 6. If iz³ + z² - z + i = 0 . Find modulus of z.

Ans:- iz³ + z² - z + i = 0
iz³ + z² + i²z + i = 0
z²(iz + 1) + i(iz + 1) = 0
(z² + i)(iz + 1) = 0
Solution are z =i & z² = - i
Considering z = i Taking modulus on both sides
|z| = |i| = 1
Now consider z² = -i, Take modulus of both sides
|z²| = |-i| = 1
|z²| = |z|² = 1
=> |z| = 1 but |z| always
Hence |z| = 1

Q.7. ……………
number z satisfying |z - i| = 4

Ans:- We know taht |z₁ -|z₂| |z₁-z₂| |z₁-z₂
Here z₁ & z₂ = 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 x⁵ - 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 z₁ & z₂ & origin form an isosceles triangle with vwetical angle .

Ans:- In this case |z₁| = |z₂|

A = |z₁|
B = |z₂|
Applying Rotation thesrem to this trianglem takin origin at O we get

Squaring both side & rearecanging we get
z₂²+ z₁² +

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 = |i²z| = |-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 & z₁ & z₂ are any two complex number prove that
( n N)

Ans:- If 1, w, w², …….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)(x² + y² xy) is a factor of g(x, y) = (x + y)ⁿ - xⁿ - yⁿ
Ans:- (xy)(x +y)(x² + y² + xy) can be writen as
f(x, y) = (xy)(x + y)(x - wy)(x - w²y)
If we can prove taht whenever f(x,y) = 0, g(x,y) = (x + y)ⁿ - xⁿ - y ⁿ is also 0, then f(x,y) is always a factor of (x + y)ⁿ - xⁿ - yⁿ
Put x = 0 which is afactor of f(x,y) in g(x,y)
we get g(0, y) = (0 + y)ⁿ - 0ⁿ - yⁿ
= yⁿ - y ⁿ = 0
i.e. (x - 0) is a factor of g(x,y)
Put y = 0 whichis a factor of f(x,y) in g(x,y)
g(x, 0) = (x + 0)ⁿ - xⁿ - 0ⁿ
= xⁿ - xⁿ = 0
i. e. (y - 0) is also a factor of g(x, y)
Put (x + y) = 0 which is a factor of f(x, y) in g(x, y)
g(x, y)/x + y = 0 = 0ⁿ - (- y)ⁿ - yⁿ
= - (- y)ⁿ - yⁿ (becaise n is 0
i.e (x + y) is also a factor of g(x, y)
Put x = wy which is afactor of f(x, y) in g(x, y)
g(wy, y) = (wy + y)ⁿ - (wy)ⁿ - yⁿ
= yⁿ [- w²ⁿ - wⁿ - 1]
g(wy, y) = - yⁿ(w²ⁿ + w + 1)
n is not a multiple of 3, n can be 3n + 1 or 3n in bot the cases w²ⁿ + wⁿ + 1 = 0, Hence g(wy, y) = 0
i.e. (x - wy) is also a factor of g(x, y)
Put x = w²y which is a factor of f(x, y) in g(x, y)
g(w²y, y) = (w²y + y)ⁿ - (w²y)ⁿ - yⁿ
= yⁿ(- w² - w²ⁿ - 1)
= - yⁿ(1 + wⁿ + w²ⁿ)
g(w²y, y) = 0
i.e. (x - w²y ) is also a factor of g(x, y) combing all the above factor which are common to both we can say xy(x + y)(x² + xy + is a factor of g(x, y) = (x + y)ⁿ - xⁿ - yⁿ

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_1/2K > 1

Ans : log_1/2K > 1 means. log_1/2K > log_1/2K > 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 z² + 4 ?

Ans:- gives remainder as i means
f(z) = K(z - 2i) + i
i.e. f(2i) = i
& f(-2i) = 2i

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