Mathematics Complex Numbers - Introduction, Algebra, Powers of i, Modulus and Conjugate

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Complex Numbers - Introduction, Algebra, Powers of i, Modulus and Conjugate

Introduction

What is the solution of `x^2 +1 =0`

What is the solution of `ax^2 + bx + c =0` when `b^2 -4ac < 0`

Let `sqrt(-1) = i`

Then `i^2 = -1`

Now, Solution `x^2 +1 =0` can be written as

`x^2 = -1 = i^2 `

`=> x = pm i`

In case of `ax^2 +bx + c =0`,

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

Generally a complex number z can written as

`z = a + ib` where `a,b in R`

`a = Re(z) \ \ \ \ b = Im (z)`

If `b = Im(z) = 0`, complex number is called purely real .

If `a = Re(z) = 0 ,` complex number is called purely imaginary .

Representation `C = {x + iy : x in R, y in R , i = sqrt(-1)}` form of set of complex numbers

Let `z_1 = a_1 + i b_1`

`z_2 = a_2 + i b_2`

then, `z_1 = z_2` iff `a_1 = a_2` and `b_1 = b_2`+

Example 1 :- If, `(x-3) + i(y -5) = 2 + 3 i`

then the value of x and y

`=> x -3 = 2 => x = 5`

`=> y -5 =3 => y =8`

Example 2 :- If `4 +i (x + y -3 ) > 2 + i (x -y -1)`

`=>{tt ((x +y =3), ( x -y=1 ) )`

`=> x =2 , y =1`

Note : Two complex number are either equal not equal. Inequality doesn't exist for the complex number.

So if `a +ib > c + id` is possible only when `b = d = 0 ` and `a > b`

Algebra of Complex Number

If `z_1 = a_1 + ib_1, z_2 = a_2 + i b_2`

then, `z_1 + z_2 = a_1 + a_2 + i (b_1 + b_2)`

and `z_1z_2 = (a_1 a_2 - b_1 b_2) + i ( a_1 b_2 + b_1 a_2)`

Ex :- If `z_1 = 3 +4 i`

`z_2 = 5 + 6 i`

then, `z_1 z_2 = (3 xx 5 - 4 xx 6) + i (20+18)`

`= -9 + 38 i`

Division of Complex Number :

Ex:- Let `z_1 = 6 + 3 i, z_2 = 2 -i`

`z_1/z_2 = (6 + 3 i)/(2 -i) xx (2 +i ) /( 2 +i)`

`= ((6 xx 2 - 3 xx 1) + i ( 6 xx 1 + 3 xx 2))/(4 +1)`

`= 1/5 ( 9 + 12 i)`

Q 2680345217

Match the terms of Column I with terms of
Column II and choose the correct option from the
codes given below.

Column I		Column II
(A)	`(1-i) * (-1 + 6 i)`	(1)	` -19/5 -(21 i)/10`
(B)	`(5 i)/(3+ 4i)`	(2)	`17/3 + i 5/3`
(C)	`(1/5 + i 2/5 ) -(4+ i 5/2)`	(3)	`5+ 7 i`
(D)	`[ (1/3 + i 7/3)+ (4+i 1/3)] - (-4/3 + i)`	(4)	` 4/5 i + 3/5 i`

(A)

`A-> 3, quad B-> 4, quad C->2, quad D->1`

(B)

`A-> 4, quad B-> 3, quad C->2, quad D->1`

(C)

`A-> 2, quad B-> 3, quad C->4, quad D->5`

(D)

`A-> 4, quad B-> 1, quad C->2, quad D->3`

Solution:

(A) `(1-i) (-1+ 6i) = -1 + 6 + i+ 6 i =5 + 7 i`

(B) `(5i)/(3+ 4i)= (5i (3-4i))/((3 +4i)(3-4i)) =(5i(3-4i))/25 =(i (3-4i))/5 = 4/5 i + 3/5 i`

(C) `(1/5 +i 2/5)-(4+i 5/2) = 17/3 + i 5/3`

(D) `[(1/3 + i 7/3 )+ (4+i 1/3)] = (-4/3 +i)] = -19/5 -(21 i)/10`

Correct Answer is `=>` (A) `A-> 3, quad B-> 4, quad C->2, quad D->1`

Power Of i

`i^3 = i^2 i = -i , i^4 = i^2.i^2 = (-1)(-1) = 1`

`i^5 = i^4i = i , i^6 = (i^3)i^3 = (-i)(-i) = -1` etc

`i^(-1) = 1/i xx i/i = i/(-1) = -i , i^(-2) = 1/i^2 = 1/(-1) = -1`

`i^(-3) = 1/i^3 = = 1/(-i)(i/i) = i/1 =i , i^(-4) = 1/i^4 = 1/1 =1`

In general, for any integer k, `i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = -1, i^(4k+3) = -i`

Here K is an Integer and not a +ve integer only

If k is -ve as , `i^(-4), i^-3, i^-2, i^-1`

same as `i^4 xx i^-4, i^4 xx i^-3, i^4 xx i^-2, i^4 xx i^-1`

same as `1, i, i^2, i^3`

`i^n + i^(n+1) + i^(n+2) + i^(n+3) = i^n(1 + i + i^2 + i^3) = i^n( 1 + i - 1 - i ) = 0`

Q 2610278119

Find the incorrect option.
(i)` i^3 = -i`
(ii) `i^4 = 1`
(iii) `i^5 =i`
(iv)`i^6 =1`

(A)

(B)

(C)

(D)

Solution:

know that,

`i = sqrt(-1) , i^2 = -1`
`i^3 =i^2 ·i = -i`
`i^4 = (i^2)^2 = 1`
`i^5 =i^2 xx i^2.i =-1 xx-1 xxi = i`
`i^6 =i^2 xx i^2 xx i^2 = -1`

Correct Answer is `=>` (D) 4

Q 1987856787

Express the complex number `(1 -i)^4` in

the form `a+ ib`.
Class 11 Exercise 5.1 Q.No. 8

Solution:

`(1 - i)^4 = [ ( 1 - i)^2]^2`

`= (1 + i^2-2 i )^2 = ( 1 - 1 - 2i)^2`

`=(-2 i )^2= 4i^2 =4(-1) =-4`

The Conjugate of a Complex Number

The conjugate of z, denoted as `barz` , is the complex number `a – ib`, i.e., `bar z = a – ib`.

Properties of conjugate :

`1. => bar(barz) = z`

Let `z = 2- 3i`

`bar z = 2 +3i `

`bar (bar z) = 2 -3i`

`2 .=> z + barz = 2Re(z)`

`z + bar z = 2 -3i + 2 + 3i= 4 = 2Re(z)`

3. `=> z - barz = 2 Im (z)`

Now, ` z -bar z = 2-3i-(2+3i) = -6i = 2Im(z)`

4 . `=> z + barz = 2-3i + 2+ 3i = 4 => z` is purely real.

5. `=> z- barz = 2-3i -( 2+ 3i) => -6i => z` is purely imaginary.

6. `=> bar(z_1 + z_2) = barz_1 + barz_2`

Now, Let ` z_1 = 3 +i` and `z_2 = 2 + i` be complex numbers. Then,

`=> bar(z_1 + z_2) =bar( 3+ i + 2+i ) = bar (5+2i ) = 5-2i `

` barz_1 + barz_2 = bar(3+i) + bar(2+i) = bar(5+2i ) = 5-2i`

7. `=> bar(z_1 · z_2) = barz_1 · barz_2`

Now let ` z_1 = 3 +i` and `z_2 = 2 - i` be complex numbers.

`=> bar(z_1 · z_2) = bar ((3 + i)( 2 - i )) = 6 +1 - 3 i + 2 i = 7 - i = 7 + i`

Now, ` barz_1 · barz_2 = (bar(3+i)).(bar(2-i)) = (3-i)(2+i) = 7 +i`

8.` => bar(z_1/z_2) = bar(z_1) / bar (z_2) `

Now `bar((z_1/z_2)) = bar(((3+i)/(2-i))) =bar((((3+i)(2+i))/((2-i)(2+i)))) = bar(1+i ) = 1 - i`

`bar(z_1) / bar (z_2) = (3-i)/(2+i) = ((3-i)(2-i)) / ((2+i)(2-i)) = 1 - i`

Q 1375278166

The conjugate of the complex number `frac{1-i}{1+i}` is

(A)

`-2i`

(B)

`-i`

(C)

`i`

(D)

`4i`

Solution:

Explanation :

`frac{1-i}{1+i} = frac{1-i}{1+i} times frac{1-i}{1-i}`

`= frac{(1-i)^{2}}{1^{2}-i^{2}}`

`= frac{1^{2}+i^{2} - 2i}{1+1}`

`= frac{1-1-2i}{2}`

`= -i`

Hence , conjugate of `frac{1-i}{1+i}` is `i` .

Correct Answer is `=>` (C) `i`

Q 1324645551

If the conjugate of a complex number is `{1}/{i-1}`, then the complex number is?

(A)

`{1}/{i-1}`

(B)

`{-1}/{i-1}`

(C)

`{1}/{i+1}`

(D)

`{-1}/{i+1}`

Solution:

If `z=x+iy` then `overline z=x-iy`

`overline z={1}/{i-1}={1}/{-1+i}`

`therefore z={1}/{-1-i}={-1}/{i+1}`

Correct Answer is `=>` (D) `{-1}/{i+1}`

The Modulus of a Complex Number

Let `z = a + ib` be a complex number. Then, the modulus of z, denoted by | z |, is defined to be the non-negative real number
`sqrt(a^2+ b^2)`

Properties of modulus :

1. `|z| =0 `

`=> z = 0, \ \ `i.e, `Re (z) =0` and `Im (z) =0`

2. `|z| = |bar z| = | -z|`

3. `- |z| <= Re (z) <= |z| ` and `- | z| <= Im (z) <= |z|`

4. `z bar z = |z|^2 , |z^2| = | bar z |^2`

If `z= 1+i`, `zbarz = (1+i)(1-i) = 2 `

`|z|^2 = |1+i|^2 = (sqrt(1^2 + 1^2))^2 = 2 = zbarz `

Now `|z^2 | = |1+i^2+2i| = |2i| = 2 `

`|barz|^2 = |1-i|^2 = 1+i^2-2i = |2i| = 2 = |z|^2`

5 . `|z_1 z_2| = |z_1| |z_2|`

Now, Let ` z_1 = 3 +i` and `z_2 = 2 - i` be complex numbers. Then,

`| z_1| = sqrt10, |z_2| = sqrt5`

`z_1 z_2 = 7 -i`

`|z_1 z_2| = sqrt50 = |z_1 | |z_2|`

6. `|z_1/z_2| = (|z_1|)/(|z_2|)` Provided `| z_2 | != 0`

Now, `|z_1 /z_2 | = | ( 3 +i )/( 2 - i ) | = | ((3+i) ( 2 + i ))/(( 2 + i ) ( 2 + i ) )|`

`= |( 6 -1 + 5 i )/(4+1) | = |( 5 + 5 i)/5|`

`5/5 | 1 +i | = sqrt2`

`(|z_1|)/( |z_2|) = sqrt10/sqrt5 = sqrt2 `

Q 2620378211

Find modulus for following complex numbers `1+2i`, `-1+2i`, `-1-2i`, `1-2i`,

Solution:

(A) `|1+2i| = sqrt(1^2 + 2^2) = sqrt5`

(B) `|-1+2i| = sqrt((-1)^2 + 2^2) = sqrt5`

(C) `|-1-2i| = sqrt((-1)^2 + (-2)^2) = sqrt5`

(D) `|1-2i| = sqrt(1^2 + (-2)^2) = sqrt5`

Q 2680445317

Find the value of |z| , If `z = (5i)/(3+ 4i) `

(A)

(B)

(C)

(D)

Solution:

By property `|z_1 /z_2| = |z_1| / |z_2|`

`|z| = |(5i)/(3+ 4i)|= ((|5i|)/(|(3 +4i)|)) = (sqrt(0^2 + 5^2))/(sqrt(3^2 + 4^2)) = 5/5 = 1`

Alternatively

`|z| = |(5i)/(3+ 4i)|= |(5i (3-4i))/((3 +4i)(3-4i))| = |(5i(3-4i))/25| = | (i (3-4i))/5| = |4/5 + 3/5 i|`

`= sqrt((4/5)^2 +(3/5)^2)=1`

Correct Answer is `=>` (C) 1

Solved Examples

Q 1927656581

Express the complex number in given equation in the form `a+ib.`

`i^(-39)`
Class 11 Exercise 5.1 Q.No. 3

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`

Q 1345678563

Find the multiplicative inverse of `2 - 3 i`

Solution:

Explanation :

Let `z=2-3i` then `overline{z} = 2+3i`

`|z|^{2} = (2)^{2}+(-3)^{2} = 13`

Therefore , the multiplicative inverse of `2 - 3 i` is given by

`z^{-1} =frac{overline{z}}{|z|^{2}} = frac{2+3i}{13} = frac{2}{13}+ i frac{3}{13}`

Alternatively :

The above working can be reproduced in the following manner also ,

`z^{-1} = frac{2+3i}{(2-3i)(2+3i)}`

`= frac{2+3i}{2^{2} - (3i)^{2}}`

`= frac{2+3i}{13}`

`=frac{2}{13}+ i frac{3}{13}`

Q 1977367286

Express the following expression in the form

of `a+ ib = ( (3+ i sqrt(5))(3-i sqrt(5)))/((sqrt(3) +i sqrt(2))-(sqrt(3) - i sqrt(2)))`
Class 11 Exercise 5.1 Q.No. 14

Solution:

We have `( (3+ i sqrt(5))(3-i sqrt(5)))/((sqrt(3) +i sqrt(2))-(sqrt(3) - i sqrt(2)))`

`= ( (3)^2 - (i sqrt(5))^2)/(sqrt(3) + isqrt(2) - sqrt(3) + i sqrt(2))`

`= (9-5i^2)/(2 sqrt(2) i) = (9-5 (-1) )/(2 sqrt(2) i) =14/(2 sqrt(2) i)`

`= 7/(sqrt(2) i) xx (sqrt(2) i)/(sqrt(2) i) = (7 sqrt(2) i)/(2i^2) = -(7 sqrt(2) i)/(2) = 0 - ((7 sqrt(2) )/2 )i`

Q 1715634560

If `(z -1)/(z+1)` is a purely imaginary number `(z != -1)`, then find the value of `| z |`.
NCERT Exemplar

Solution:

Let `z = x + iy`

` (z -1)/(z+1) = (x + iy - 1)/(x + iy + 1) , z != -1`

` = (x - 1 + iy )/(x + 1 + iy) = ( (x - 1 + iy ) (x + 1 - iy ))/( (x + 1 + iy ) (x + 1 - iy ))`

` = ( (x^2 - 1 ) + iy (x + 1)- iy (x -1)- i^2y^2)/( (x + 1)^2 - (iy)^2)`

` => (z -1)/(z+1) = ( (x^2 -1) + y^2 + i [y(x + 1) - y(x -1)])/((x + 1)^2 + y^2)`

Given that, `(z -1)/(z+1)` is a purely imaginary numbers.

Then, ` ( (x^2 - 1)+ y^2)/( (x^2 + 1)^2 + y^2) = 0`

` => x^2 - 1 + y^2 = 0 => x^2 + y^2 = 1`

` => sqrt(x^2 + y^2) = sqrt(1) => |z| = 1 quad [∵ |z| = sqrt(x^2 + y^2) ]`

Q 1775045866

What is the conjugate of `(2-i)/(1-2i)^2 `?

NCERT Exemplar

Solution:

Given that, ` z = (2-i)/(1-2i)^2 = ( 2-i) /( 1 + 4i^2 - 4i)`

` = (2-i)/( 1- 4 - 4i) = (2-i)/(-3 -4i)`

` = (2-i)/(-3 -4i) = - [(((2-i)(3-4i))/((3+4i)(3-4i)))]`

` = - ( (9- 8i -3i + 4i^2 )/(9 + 16)) = - ( -11i +2)/(25)`

` = - (-1)/(25) (2 - 11 i) => z = 1/(25) (-2 - 11 i)`

`:. bar z = 1/(25) (-2 - 11 i) = (-2)/(25) - (11)/(25) i`

Q 1725656561

If `z` is a complex number, then prove `|z^2| = | z |^2 `
NCERT Exemplar

Solution:

If `z` is a complex number, then `z = x + iy`

` | z | =| x + iy|` and `|z|^ 2 = |x + iy|^ 2`

` => |z|^2 = x^2 + y^2`........(1)

and `z^2 = (x + iy)^2 = x^2 + i^2y^2 + i2xy`

`z^2 = x^2 - y^2 + i2xy`

`=> | z^2 | = sqrt((x^2 - y^2 )^2 + (2xy)^2 )`

`=> | z^2| = sqrt( x^4 + y^2 - 2x^2 y^2 + 4 x^2y^2)`

`=> | z^2| = sqrt( x^4 + y^2 + 2x^2 y^2) = sqrt((x^2 + y^2)^2)`

`=> | z^2| = x^2 + y^2` .........(2)

From Eqs. (1) and (2),

` | z|^2 = |z^2 |`

Q 1725245161

If `| z + 4| <= 3`, then the greatest and least values of `| z + 1|` are ...... and ..........
NCERT Exemplar

Solution:

Given that, `| z + 4 | <= 3`

For the greatest value of `| z + 1|`.

`=> | z+1| = |z +4 -3 | <= |z+4| +| -3|`

` = | z + 4 -3 | <= 3 + 3`

` = | z + 4 - 3 | <= 6`

So, greatest value of `| z + 1| = 6`

For, now, least value of `| z + 1|`. we know that the least value of the modulus of a complex

number is zero. So, the east value of `| z + 1|` is zero.

Q 1785156067

Where does `z` lie, if ` | ( z - 5i)/(z + 5i) | = 1`?
NCERT Exemplar

Solution:

Let `z =x + iy`

Given that, ` | ( z - 5i)/(z + 5i) | = | ( x + iy - 5i) /( x + iy + 5i) |`

` => | ( z - 5i)/(z + 5i) | = (|x + i (y- 5)|)/(| y + i (y + 5)|)`

`=> | ( z - 5i)/(z + 5i) | = sqrt( x^2 + (y - 5)^2)/sqrt( x^2 + (y + 5)^2)`

On squaring both sides, we get

` x^2 + (y- 5)^2 = x^2 + (y + 5)^2`

` => -10y = + 10y`

` => 20y = 0`

` :. y = 0`

So, `z` lies on real axis.

Q 2036767672

The locus represented by `| z -1| =| z- i |` is a line perpendicular to the
join of the points `(1, 0)` and `(0, 1)`.
NCERT Exemplar

(A)

True

(B)

False

Solution:

Let `z = x + iy`

`| z -1| = | z - i |`

Then `| x - 1 + iy | = | x - i (1 - y) |`

` (x -1)^2 + y^2 = x^2 + (1- y)^2`

` x^2 - 2x + 1 + y^2 = x^2 + 1 + y^2 - 2y`

` - 2x + 1 = 1 - 2y`

` - 2x + 2y = 0`

` x - y = 0`

Equation of a line through the points `(1, 0)` and `(0, 1)`,

` y - 0 = (1 -0)/(0-1) (x -1)`

`=> y = - (x - 1) => x + y = 1`

which is perpendicular to the line `x - y = 0` .

Correct Answer is `=>` (A)