Basics & Formulas Ch-4 Class11 | Complex Numbers
Complex Numbers Chapter-4 Class-11
Following concepts will be discussed in this Blog
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All rational and irrational numbers are called real numbers.
The numbers which are not real are called non real numbers. These numbers are also called imaginary numbers.
Imaginary numbers :-
Square root of a negative real number is called the imaginary number.
Complex Number:
All real and non- real numbers are called complex number
i = √-1 , i2 = -1, i3 = - i, i4 = 1
If Y and Z are two complex numbers such that
Y = a + ib and Z = c + id,
Then Y = Z ⇒ a = c and b = d
Addition of two complex numbers
Two complex numbers are added by simply adding their real and imaginary parts
If Y = a + ib and Z = c + id then
Y + Z = (a + c) + i(b + d)
Properties of addition of two complex numbers
i) Addition of two complex numbers holds the closure property. This means that addition of two complex numbers is also a complex number.
ii) Addition of complex numbers is commutative.
⇒ Y + Z = Z + Y
iii) Addition of complex numbers holds associative law
⇒ (X + Y) + Z = X + (Y + Z)
iv)Existence of additive identity:
0 is the additive identity for the
addition of complex number.
∵
Z + 0 = Z
v) Existence of additive inverse :
If Z is a complex number then - Z is called the additive inverse of Z.
Subtraction of two complex numbers :
Two complex numbers are subtracted by subtracting their corresponding real and imaginary parts.
If Y = a + ib, Z = c + id then
Y - Z = (a + ib) - (c + id)
= (a - c) + i(b - d)
Properties of
Multiplication of two complex numbers
i) Closure property
:
Product of two complex numbers is also a complex number.
ii) Commutative law :
Multiplication of complex
numbers is commutative. ⇒
YZ = ZY
iii) Associative Law:
Multiplication of complex numbers is associative. ⇒ (XY)Z = X(YZ)
iv) Existence of multiplicative identity:
1 is called the multiplicative identity for the product of complex number.
v) Multiplicative inverse :
If Z is a complex number then Z-1 is called the inverse of Z. ∵ Z x Z-1 = 1
vi) Distributive law :
Product of complex numbers holds the distributive property. ⇒ X(Y + Z) = XY + XZ
Conjugate of a complex number
If Z= a+ib is a complex number then conjugate of Z is denoted by and is given by
Note:- Conjugate of the conjugate of complex number is the complex number itself.
If Z = a + ib, then
Modulus of complex number
If Z = a + ib is a complex number then modulus
of Z is denoted by |Z| and is given by
Argand Plane :
The plane having a complex number assigned to each of its point is called a complex plan or Argand plane.
Every point of the form (x, y) can be represented in the cartesian coordinate plane.
When a complex number is represented in the plan then the plane is called a complex plane or Argand plane.
In Argand plane x- axis is called
real axis and y-axis is called imaginary axis.
If Z = x + iy then
Putting x = rcosθ and y = rsinθ in Z = x + iy we get
Z = rcosθ + irsinθ or Z = r(cosθ + isinθ)
This is called polar form of a complex number.
If Z = a + ib, then to convert it into polar form we put a = rcosθ and b = rsinθ
tanθ = b/a and θ = tan-1(b/a)
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If (a, b) lie in first quadrant then Argument = θ If (a, b) lie in second quadrant then Argument = π - θ If (a, b) lie in third quadrant then Argument = -π + θ If (a, b) lie in fourth quadrant then Argument = - θ If complex number Z is purely +ve real number then Argument is 0 If complex number Z is purely -ve real number then Argument is π If complex number Z is purely +ve imaginary number then Argument is π/2 If complex number Z is purely -ve imaginary number then Argument is - π/2 |
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Method of finding the square root of a complex number This can be
explained by taking an example \[Find\; the\;
square \; root\; of\; the\; complex\; number\; \; z=\sqrt{3+i4}\]\[Let\; \;
a+ib=\sqrt{3+i4}\]Squaring on both side we
get,\[(a)^{2}+(ib)^{2}+2iab=3+i4\]\[(a)^{2}-b^{2}+2abi=3+i4\]Compairing real
and imaginary parts we ger, \[(a)^{2}-b^{2}=3 .......(1),\: \: and\:
2ab=4\]\[Now\: using\: the\: formula\: \:
(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2}+(2ab)^{2}\]\[(a^{2}+b^{2})^{2}=(3)^{2}+(4)^{2}=9+16=25\]\[\Rightarrow
(a^{2}+b^{2})=5 ......(2)\]Adding eqn(1) and eqn(2) we
get\[2a^{2}=8\Rightarrow a^{2}=4\Rightarrow a=\pm 2\]From eqn(1) - eqn(2) we
get\[2b^{2}=2\Rightarrow b^{2}=1\Rightarrow b=\pm 1\]\[\sqrt{3+i4}=a+ib=\pm
2\pm 1i=\pm (2+i)\]Here between 2 and i we apply +ve sign because 2ab is
positive. If the value of 2ab is -ve then between a and b we apply -ve sign
in the square root of the complex number. |
Squaring on both sides we get\[(x+5)^{2}=\left [ 2\sqrt{-4} \; \right ]^{2}\]\[x^{2}+5^{2}+2\times x\times 5=4\times (-4)\]\[x^{2}+25+10x+16=0\]\[x^{2}+10x+41=0\]
Now divide the given equation with this equation we get
Dividend = Divisor X Quotient + RemainderP(x) = (x2 + 10x + 41)(x2 – x + 4) – 160= 0 X (x2 – x + 4) – 160 = - 160 Ans







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