Mathematics Argand Plane , Polar Representation and Argument of Complex Number

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Argand Plane , Polar Representation and Argument of Complex Number

Ordered Pair :

`z= x + iy`

`z=(x,y )` ordered Pair

`x= Re (z), y = Im (z)`

We associate a one to one corresponding behaviour between `z=x+iy` and the point `(x,y)` in `x-y ` plane .

Argand Diagram :

The complex number `(x+ iy)` which correspond to the ordered pair `(x,y)` can be represented

`x-` axis `-> ` Real axis

`y-` axis `-> ` Imaginary axis

Definition : The Plane having a complex number assigned to each of its pair is called Argand Plane , (Also presented as complex plane or `z-` plane).

Q 2600356218

Plot the complex numbers `z_1 = 2 + 3i, z_2 = −3 + 2i, z_3 = −3 − 2i, z_4 = 2 − 5i, z_5 = 6, z_6 = i` on an Argand diagram.

Solution:

Note:
(i) Purely real numbers lie on the real axis. Purely imaginary numbers lie on the imaginary axis.
(ii) Another observation is that complex conjugate pairs (such as `−3 + 2i` and `−3 − 2i` ) lie symmetrically about the x axis.

Representation of `|z|` in Argand Plane and its meaning :

`|z|= sqrt(x^2+y^2) =OP`

Representation of `bar z` (conjugate of `z`) as Argand Plane

`Q` is the conjugate of `P`

If `P= (x,y)` then `Q=(x, -y)`

Q 2680156017

Represent the `|z|` in Argand Plane for following points A(3+4i) and B(3-4i)

Solution:

`|A| = sqrt(3^2 + 4^2) = 5`

`|B| = sqrt(3^2 + 4^2) = 5`

Polar Representation :

`Z= (r, theta)` in Polar representation of point `P(z=x+iy)`

where, `x= r cos theta`

`y= r sin theta`

and `- pi < theta le pi`

`z= x+iy= r (cos theta + i sin theta)`

`= r cos theta + r sin theta `

`:.` Polar form is also called as trigonometrical form

Q 2610356219

Express the complex number in polar form `5+2i`

Solution:

So, first find the absolute value of r

`r = |z| = sqrt(a^2 + b^2) = sqrt(5^2+2^2)`

`=5.39 `

Now find the argument `θ`

Since `a>0` , use the formula `θ= tan^(-1)(b/a)`

`θ= tan^(-1)(2/5) = 0.38` (θ is measured in radians.)

Therefore, the polar form of `5 + 2 i` is about `5.39(cos(0.38)+isin(0.38))`

Argument of Complex Number :

`Arg (z)= theta`

Arguments are of 3 types,

(i) General Arguments
(ii) Principal value of Argument or Amplitude or Argument
(iii) Least positive Argument

General Argument :

General value of Argument of `z= 2 n pi + theta, n in I`

Ex- Let `z= 1+i`

`=> (x,y)=(1,1)`

`=>` General value of Argument `= 2 n pi + pi/4`

Put `n =-1 `

One value of `theta = -2 pi + pi/4 =(-7 pi)/4`

`theta = (-15 pi)/4 ,(-7 pi)/4 , pi/4 , (9 pi)/4 , (17 pi)/4`

Principal value of Argument on Amplitude or Argument :

The unique value of e such that `- pi < theta le pi` is called principal value of argument.

Mathematically to find Principal Argument of `z`

Let `z = a + ib`

Find `alpha= tan^(-1) ((|b|)/(|a|))`

Case I : If `z` lies in `I` quadrant i.e. `a, b > 0` then `amp (z) = theta = alpha`.

Case II : lf `z` lies in `II` quadrant i.e. `a < 0, b > 0` then `amp(z) = theta = (pi - alpha)`

Case III : If `z` lies in Ill quadrant i.e. `a < 0, b < 0` then `amp(z) = theta =- (pi - alpha)`

Case IV : lf `z` lies in IV quadrant i.e `a > 0, b < 0` then `amp(z) = theta =- alpha`.

Q 2670356216

Find the principal argument for following points `A(1+ sqrt3 i) , B(-1 + sqrt 3 i), C(-1 -sqrt 3 i), D(1- sqrt 3 i)`

Solution:

`A (1+ sqrt 3 i)` Here `, a>0 , b > 0` `A` lies in `I` quadrant

`amp (A)= theta_A =tan^(-1) ((|b|)/(|a|))`

`=tan^(-1) |(sqrt 3)/1|= pi/3`

For ` B(-1 + sqrt 3 i) ,` `a < 0, b > 0` then B lies in `II` quadrant

`amp (B)= theta_B= pi - tan^(-1) ((| sqrt 3|)/(|-1|))`

`=pi - pi/3 =(2 pi)/3`

For `C(-1 -sqrt 3 i)`, `a < 0, b < 0` ,`C` lies in Ill quadrant

`amp (C)= theta_C= -pi + tan^(-1) ((|sqrt 3|)/(|-1|))= -pi + pi/3`

`=-(2 pi)/3`

For `D(1- sqrt 3 i)` `a > 0, b < 0` `D` lies in IV quadrant

`amp (D) = theta_D =-tan^(-1) ((|sqrt 3|)/(|-1|))`

`=- pi/3`

Least Positive Value of Argument :

The value of `theta` such that ` 0 < theta < 2 pi` is called the Least Positive Argument.

Quadratic Equation with discriminant ` < 0`

`ax^2 +bx + c=0` `a,b, c in R` and `a ne 0`

`b^2-4ac < 0`

Then, `-(b^2-4ac) > 0`

`x= (-b pm sqrt(b^2-4ac))/(2a)`

`=(-b pm i sqrt(4ac -b^2))/(2a)`

Complex roots always occur in pairs.

Q 1907167988

Solve the following equation:

`sqrt(2) x^2 +x +sqrt(2) =0`
Class 11 Exercise 5.3 Q.No. 7

Solution:

`sqrt(2) x^2 +x +sqrt(2) =0`

`a =sqrt(2) , b=1, c=sqrt(2)`

`x= (-b pm sqrt (b^2 -4ac))/(2a)`

`x= (-1 pm sqrt ((1)^2 -4 xx sqrt (2) xx sqrt(2) ))/(2 sqrt(2) )`

`x= (-1 pm sqrt (1-8))/(2 sqrt(2))`

`x= (-1 pm sqrt(7) i)/(2 sqrt(2))`

Solved Examples

Examples:

Q 2017301280

Find the principal argument of `(1 + i sqrt(3))^2`.

NCERT Exemplar

Solution:

Given that, ` z = (1 + isqrt(3))^2`

` => z = 1- 3 + 2i sqrt(3) => z = - 2 + i2 sqrt(3)`

` => tan alpha = |(2sqrt(3))/(-2)| = | - sqrt(3) | = sqrt(3)`

` => tan alpha = tan (pi/3) => alpha = pi/3`

` ∵ Re (z) < 0` and `Im (z) > 0`

`=> arg (z) = pi - pi/3 => = (2pi)/3`

Q 1917467389

Convert `(1 -i)` in the polar form.
Class 11 Exercise 5.2 Q.No. 3

Solution:

We have `1 - i = r (cos theta + i sin theta)`

`=> r cos theta = 1, r sin theta =-1`

By squaring and adding, we get

`r^2 (cos^(2) theta + sin^(2) theta ) =1^2 +(-1)^2`

`=> r^2 *1 = 1 + 1 => r^2 = 2`

` :. r = sqrt(2) ` , By dividing `(r sin theta)/(r cos theta) =-1/1 =-1`

`=> tan theta = -1 ` i.e., `theta` lies in fourth quadrant.

`=> theta = -45`

`=> theta = -pi/4`

`:. ` Polar form of `1-i`

`= sqrt(2) (cos (-pi/4) + i sin (-pi/4) )`

Q 1967667585

Convert `(-1 -i)` in the polar form.
Class 11 Exercise 5.2 Q.No. 5

Solution:

we have `-1 - i = r (cos theta + i sin theta)`

`=> r cos theta -1` and `r sin theta = -1`

By squaring and adding. we get

`r^2 (cos^(2) theta + sin^(2) theta) = (-1)^2 + (-1)^2`

`=> r^2 * 1 =1+1`

`=> r^2 =2`

`:. r =sqrt(2)`

By dividing `(r sin theta)/(r cos theta) = -1/-1 =1 => tan theta =1`

`:. theta` lies in `III^(rd)` quadrant.

`theta = -180^(circ) + 45^(circ) = -135^(circ) ` or `theta = -(3 pi)/4`

`:.` Polar form of `-1 -i`

`= sqrt(2) (cos (-(3 pi)/4) + i sin (- (3 pi)/4))`

Q 1927178081

Solve the following equation:

`sqrt(3) x^2 - sqrt(2) x + 3 sqrt(3) =0`
Class 11 Exercise 5.3 Q.No. 8

Solution:

We have `a= sqrt(3) , b = -sqrt(2) , c= 3 sqrt(3)`

`:.` Discriminant of the equation is `D = b^2 -4ac`

`= (- sqrt(2) )^2 -4 xx sqrt(3) xx 3sqrt(3)`

`=>D =2 -36 = -34`

`:. x = (-b pm sqrt(D) )/(2a) = (sqrt(2) pm sqrt (-34))/(2 xx sqrt(3)) = (sqrt(2) pm sqrt(34) i)/(2 sqrt(3) )`

`:. x = (sqrt(2) pm sqrt(34) i)/(2 sqrt(3))`

Q 2643567443

Convert the complex numbers `i^(-39)` in the polar form.

Solution:

`i^(-39)= (i^2)^(-19) i^(-1) = (-1)^(-19)* i^(-1)`

`= 1/(-1)^(19) xx 1/i = -1/i xx i/i`

`= -i/(i^2) = -i/-i =i`

Now

`z =i = r (cos theta + i sin theta)`

`:. r cos theta = 0 , r sin theta =1`

Squaring and adding `r^2 =1, :. r =1`

Now `sin theta = 1, cos theta = 0` at `theta =pi/2`

`:.` Polar form of `z` is `cos( pi/2) + i sin (pi/2)`

Q 1745534463

If `arg (z -1) = arg (z +3i)`, then find `x -1 : y`, where `z = x + iy`.
NCERT Exemplar

Solution:

Given that, `arg( z - 1 ) = arg ( z + 3i )`

and let `z = x + iy`

Now, `arg (z- 1) = arg (z + 3i)`

` => arg (x + iy - 1) = arg (x + iy + 3i)`

` => arg (x- 1 + iy) = arg [x + i (y + 3)]`

`=> tan^(-1) (y/(x-1)) = tan^(-1) ((y + 3)/x)`

` => y/(x-1) = (y+3)/x => xy = (x - 1) (y + 3)`

`=> xy = xy - y + 3x - 3 => 3x - 3 = y `

` => (3(x - 1))/y = 1 => (x -1)/y = 1/3`

`:. (x -1) : y = 1 : 3`

Q 1775623566

If `a = cos theta + i sin theta`, then find the value of ` ( 1+ a)/(1-a)`
NCERT Exemplar

Solution:

Given that. `a = cos theta + i sin theta`

`:. ( 1+ a)/(1-a) = ( 1 + cos theta + i sin theta)/( 1 - cos theta - i sin theta)`

` = ( 1 + 2cos^2 theta//2 -1 + 2i sin theta//2 . cos theta//2) /(1 - 1+ 2sin^2 + theta//2 - 2isin theta//2 .cos theta//2) `

` = ( 2cos theta//2(cos theta//2 + i sin theta//2))/(2sin theta//2 (sin theta//2 - icos theta//2))`

` = -(2cos theta//2 ( cos theta//2 + i sin theta//2))/(2i sin theta//2 + cos theta//2 + i sin theta//2) = - 1/i cot theta//2`

` = (+i^2)/i cot theta//2 = icot theta//2 .`