`star` Introduction

`star` Complex Numbers

`star` Addition of two complex numbers

`star` Difference of two complex numbers

`star` Multiplication of two complex numbers

`star` Division of two complex numbers

`star` Power of i

`star` The square roots of a negative real number

`star` Identities

`star` The Modulus and the Conjugate of a Complex Number

`star` Complex Numbers

`star` Addition of two complex numbers

`star` Difference of two complex numbers

`star` Multiplication of two complex numbers

`star` Division of two complex numbers

`star` Power of i

`star` The square roots of a negative real number

`star` Identities

`star` The Modulus and the Conjugate of a Complex Number

● We have seen that the equation `color(blue)(x^2 + 1 = 0)` has no real solution as `x^2 + 1 = 0` gives `x^2 = – 1` and square of every real number is non-negative.

● So, we need to extend the real number system to a larger system so that we can find the solution of the equation `x^2 = – 1.`

● In fact, the main objective is to solve the equation `color(blue)(ax^2 + bx + c = 0),` where `color(blue)(D = b^2 – 4ac < 0),` which is not possible in the system of real numbers.

● So, we need to extend the real number system to a larger system so that we can find the solution of the equation `x^2 = – 1.`

● In fact, the main objective is to solve the equation `color(blue)(ax^2 + bx + c = 0),` where `color(blue)(D = b^2 – 4ac < 0),` which is not possible in the system of real numbers.

● Let us denote `color(red)(sqrt (−1))` by the symbol `color(red)(i).` Then, we have `color(blue)(i^2 = −1) `. This means that `i` is a solution of the equation `x^2 + 1 = 0.`

● A number of the form `a + ib,` where `a` and `b` are real numbers, is defined to be a complex number.

`color(red)("For example")`, `2 + i3, (– 1) + i sqrt(3)` are `color(blue)("complex numbers.")`

● For the complex number `color(blue)(z = a + ib)`,

`color(blue)(a)` is called the `color(blue)("real part")`, denoted by `color(blue)("Re z)` and `color(blue)(b)` is called the `color(blue)("imaginary part")` denoted by `color(blue)(Im z)` of the complex number `z.`

For example, if `z = 2 + i5,` then `Re z = 2` and `Im z = 5.`

● Two complex numbers `z_1 = a + ib` and `z_2 = c + id` are equal if `a = c` and `b = d.`

● A number of the form `a + ib,` where `a` and `b` are real numbers, is defined to be a complex number.

`color(red)("For example")`, `2 + i3, (– 1) + i sqrt(3)` are `color(blue)("complex numbers.")`

● For the complex number `color(blue)(z = a + ib)`,

`color(blue)(a)` is called the `color(blue)("real part")`, denoted by `color(blue)("Re z)` and `color(blue)(b)` is called the `color(blue)("imaginary part")` denoted by `color(blue)(Im z)` of the complex number `z.`

For example, if `z = 2 + i5,` then `Re z = 2` and `Im z = 5.`

● Two complex numbers `z_1 = a + ib` and `z_2 = c + id` are equal if `a = c` and `b = d.`

Q 3028801701

If `4x + i(3x – y) = 3 + i (– 6),` where x and y are real numbers, then find the values of x and y.

We have

`4x + i (3x – y) = 3 + i (–6)` ... (1)

Equating the real and the imaginary parts of (1), we get

`4x = 3, 3x – y = – 6,`

which, on solving simultaneously, give `x = 4/3 ` and `y = 33/4`

Let `z_1 = a + ib` and `z_2 = c + id` be any two complex numbers.

● Then, the sum `z_1 + z_2` is defined as follows: `z_1 + z_2 = (a + c) + i (b + d),` which is again a complex number.

For example, `(2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8`

● The addition of complex numbers satisfy the following properties:

`color(blue)("(i) The closure law")` The sum of two complex numbers is a complex number,

i.e., `z_1 + z_2` is a complex number for all complex numbers `z_1` and `z_2.`

`color(blue)("(ii) The commutative law")` For any two complex numbers `z_1` and `z_2,`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)(z_1 + z_2 = z_2 + z_1)`

`color(blue)( "(iii) The associative law")` For any three complex numbers `z_1, z_2, z_3,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)).`

`color(blue)("(iv) The existence of additive identity ")` There exists the complex number `0 + i 0` (denoted as `0`), called the additive identity or the zero complex number, such that, for every complex number z, `color(red)(z + 0 = z.)`

`color(blue)"(v) The existence of additive inverse"` To every complex number `z = a + ib,` we have the complex number `– a + i(– b)` (denoted as – z), called the additive inverse or negative of `z.` We observe that `z + (–z) = 0` (the additive identity).

● Then, the sum `z_1 + z_2` is defined as follows: `z_1 + z_2 = (a + c) + i (b + d),` which is again a complex number.

For example, `(2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8`

● The addition of complex numbers satisfy the following properties:

`color(blue)("(i) The closure law")` The sum of two complex numbers is a complex number,

i.e., `z_1 + z_2` is a complex number for all complex numbers `z_1` and `z_2.`

`color(blue)("(ii) The commutative law")` For any two complex numbers `z_1` and `z_2,`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)(z_1 + z_2 = z_2 + z_1)`

`color(blue)( "(iii) The associative law")` For any three complex numbers `z_1, z_2, z_3,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)).`

`color(blue)("(iv) The existence of additive identity ")` There exists the complex number `0 + i 0` (denoted as `0`), called the additive identity or the zero complex number, such that, for every complex number z, `color(red)(z + 0 = z.)`

`color(blue)"(v) The existence of additive inverse"` To every complex number `z = a + ib,` we have the complex number `– a + i(– b)` (denoted as – z), called the additive inverse or negative of `z.` We observe that `z + (–z) = 0` (the additive identity).

● Given any two complex numbers `z_1` and `z_2,` the difference `z_1 – z_2` is defined as follows:

`color(red)(z_1 – z_2 = z_1 + (– z_2).)`

`color(red)(z_1 – z_2 = z_1 + (– z_2).)`

Let `z_1 = a + ib` and `z_2 = c + id` be any two complex numbers.

● Then, the product `z_1 z_2` is defined as follows: `z_1 z_2 = (ac – bd) + i(ad + bc)`

For example, `(3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28`

● The multiplication of complex numbers possesses the `color(red)("following properties")`,

`color(blue)"(i) The closure law"` The product of two complex numbers is a complex number, the product `z_1 z_2` is a complex number for all complex numbers `z_1` and `z_2.`

`color(blue)"(ii) The commutative law"` For any two complex numbers `z_1` and `z_2,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){z_1 z_2 = z_2 z_1}`

.

`color(blue)"(iii) The associative law"` For any three complex numbers `z_1, z_2, z_3,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){ (z_1 z_2) z_3 = z_1 (z_2 z_3)}.`

`color(blue)"(iv) The existence of multiplicative identity"` There exists the complex number `1 + i 0` (denoted as 1), called the multiplicative identity such that `color(red)(z.1 = z),` for every complex number `z.`

`color(blue)"(v) The existence of multiplicative inverse"` For every non-zero complex number `z = a + ib` or `a + bi(a ≠ 0, b ≠ 0),`

we have the complex number `a/(a^2+b^2)+i(-b)/(a^2+b^2)` (denoted by `1/z` or `z^(-1)` called the multiplicative inverse of `z`

such that `z. 1/z = 1 ` (the multiplicative identity).

`color(blue) "(vi) The distributive law"` For any three complex numbers `z_1, z_2, z_3,`

`(a) \ \ color(red)(z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3)`

`(b) \ \ color(red)((z_1 + z_2) z_3 = z_1 z_3 + z_2 z_3)`

● Then, the product `z_1 z_2` is defined as follows: `z_1 z_2 = (ac – bd) + i(ad + bc)`

For example, `(3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28`

● The multiplication of complex numbers possesses the `color(red)("following properties")`,

`color(blue)"(i) The closure law"` The product of two complex numbers is a complex number, the product `z_1 z_2` is a complex number for all complex numbers `z_1` and `z_2.`

`color(blue)"(ii) The commutative law"` For any two complex numbers `z_1` and `z_2,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){z_1 z_2 = z_2 z_1}`

.

`color(blue)"(iii) The associative law"` For any three complex numbers `z_1, z_2, z_3,`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){ (z_1 z_2) z_3 = z_1 (z_2 z_3)}.`

`color(blue)"(iv) The existence of multiplicative identity"` There exists the complex number `1 + i 0` (denoted as 1), called the multiplicative identity such that `color(red)(z.1 = z),` for every complex number `z.`

`color(blue)"(v) The existence of multiplicative inverse"` For every non-zero complex number `z = a + ib` or `a + bi(a ≠ 0, b ≠ 0),`

we have the complex number `a/(a^2+b^2)+i(-b)/(a^2+b^2)` (denoted by `1/z` or `z^(-1)` called the multiplicative inverse of `z`

such that `z. 1/z = 1 ` (the multiplicative identity).

`color(blue) "(vi) The distributive law"` For any three complex numbers `z_1, z_2, z_3,`

`(a) \ \ color(red)(z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3)`

`(b) \ \ color(red)((z_1 + z_2) z_3 = z_1 z_3 + z_2 z_3)`

Given any two complex numbers `z_1` and `z_2,`

where `z_2 ≠ 0 ,` the quotient `z_1/z_2` is defined by ` color(red)(z_1/z_2=z_1* 1/z_2)`

where `z_2 ≠ 0 ,` the quotient `z_1/z_2` is defined by ` color(red)(z_1/z_2=z_1* 1/z_2)`

we know that

● `color(blue)(i^3) = i^2i = (−1) i = −i , `

● ` color(blue)(i^4) = (i^2)^2 = (−1)^2 =1 `

● ` color(blue)(i^5) = (i^2)^2* i = (−1 )^2 i = i , `

● `color(blue)(i^6) = (i^2)^3 = (−1)^3 =-1` , etc

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

● `color(blue)(i^3) = i^2i = (−1) i = −i , `

● ` color(blue)(i^4) = (i^2)^2 = (−1)^2 =1 `

● ` color(blue)(i^5) = (i^2)^2* i = (−1 )^2 i = i , `

● `color(blue)(i^6) = (i^2)^3 = (−1)^3 =-1` , etc

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

● Note that `color(blue)(i^2 = –1)` and `color(blue)(( – i)^2) = i^2 = – 1`

Therefore, the square roots of `– 1` are `i, – i.`

● However, by the symbol ` sqrt(−1)` , we would mean `i` only.

Now, we can see that `color(blue)(i)` and `color(blue)(–i)` both are the solutions of the equation `color(blue)(x^2 + 1 = 0)` or `x^2 = –1.`

● Similarly

`color(blue)((sqrt3 i)^2) = (sqrt3)^2 i^2 = 3(-1) = -3`

`color(blue)((-sqrt3 i)^2) = (-sqrt3)^2 i^2 = -3`

The square roots of `color(red)(–3)` are `color(red)(sqrt (3) * i)` and `color(red)(− sqrt(3) * i)`

● Generally, if `a` is a positive real number, `sqrt(−a) = sqrt(a)* sqrt(−1 )= sqrt(a)* i ,.`

We already know that `color(blue)(sqrt(a) ×sqrt( b) = sqrt(ab))` for all positive real number `a` and `b.`

This result also holds true when either `a > 0, b < 0` or `a < 0, b > 0.`

● What if `color(red)(a < 0, b < 0 \ \?) ` Let us examine.

Note that

`i^2 = sqrt(−1) * sqrt(−1) = sqrt((−1) (−1)) ` (by assuming `a × b = ab` for all real numbers) `= sqrt(1) = 1,` which is a contradiction to the fact that `i ^2 = − 1.`

● Therefore, `color(red)(sqrt(a) × sqrt(b) != sqrt(ab))` if both `a` and `b` are negative real numbers.

Further, if any of `a` and `b` is zero, then, clearly, `sqrt(a) × sqrt(b) = sqrt(ab) = 0.`

Therefore, the square roots of `– 1` are `i, – i.`

● However, by the symbol ` sqrt(−1)` , we would mean `i` only.

Now, we can see that `color(blue)(i)` and `color(blue)(–i)` both are the solutions of the equation `color(blue)(x^2 + 1 = 0)` or `x^2 = –1.`

● Similarly

`color(blue)((sqrt3 i)^2) = (sqrt3)^2 i^2 = 3(-1) = -3`

`color(blue)((-sqrt3 i)^2) = (-sqrt3)^2 i^2 = -3`

The square roots of `color(red)(–3)` are `color(red)(sqrt (3) * i)` and `color(red)(− sqrt(3) * i)`

● Generally, if `a` is a positive real number, `sqrt(−a) = sqrt(a)* sqrt(−1 )= sqrt(a)* i ,.`

We already know that `color(blue)(sqrt(a) ×sqrt( b) = sqrt(ab))` for all positive real number `a` and `b.`

This result also holds true when either `a > 0, b < 0` or `a < 0, b > 0.`

● What if `color(red)(a < 0, b < 0 \ \?) ` Let us examine.

Note that

`i^2 = sqrt(−1) * sqrt(−1) = sqrt((−1) (−1)) ` (by assuming `a × b = ab` for all real numbers) `= sqrt(1) = 1,` which is a contradiction to the fact that `i ^2 = − 1.`

● Therefore, `color(red)(sqrt(a) × sqrt(b) != sqrt(ab))` if both `a` and `b` are negative real numbers.

Further, if any of `a` and `b` is zero, then, clearly, `sqrt(a) × sqrt(b) = sqrt(ab) = 0.`

● We prove the following identity

`color(blue)((z_1 + z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2 ,)` for all complex numbers `z_1` and `z_2.`

`color(red)"Proof"` We have, `(z_1 + z_2)^2 = (z_1 + z_2) * (z_1 + z_2),`

`= (z_1 + z_2) z_1 + (z_1 + z_2) z_2` (Distributive law)

`= z_1^2 + z_2z_1 + z_1z_2 + z_2^2` (Distributive law)

`= z_1^2 + z_2z_1 + z_1z_2 + z_2^2` (Commutative law of multiplication )

`=z_1^2 + z_2^2 + 2z_1z_2 ,`

● Similarly, we can prove the following identities:

(i) `color(blue)((z_1 - z_2)^2 = z_1^2 + z_2^2 - 2z_1z_2)`

(ii) `color(blue)(( z_1 + z_2)^3 = z_1^3 + 3z_1^2 z_2 + 3z_1z_2^2 + z_2^3)`

(ii) `color(blue)(( z_1 - z_2)^3 = z_1^3 - 3z_1^2 z_2 + 3z_1z_2^2 - z_2^3)`

(iv) `color(blue)(z_1^2 – z_2^2 = (z_1 + z_2) *( z_1 – z_2))`

`color(blue)((z_1 + z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2 ,)` for all complex numbers `z_1` and `z_2.`

`color(red)"Proof"` We have, `(z_1 + z_2)^2 = (z_1 + z_2) * (z_1 + z_2),`

`= (z_1 + z_2) z_1 + (z_1 + z_2) z_2` (Distributive law)

`= z_1^2 + z_2z_1 + z_1z_2 + z_2^2` (Distributive law)

`= z_1^2 + z_2z_1 + z_1z_2 + z_2^2` (Commutative law of multiplication )

`=z_1^2 + z_2^2 + 2z_1z_2 ,`

● Similarly, we can prove the following identities:

(i) `color(blue)((z_1 - z_2)^2 = z_1^2 + z_2^2 - 2z_1z_2)`

(ii) `color(blue)(( z_1 + z_2)^3 = z_1^3 + 3z_1^2 z_2 + 3z_1z_2^2 + z_2^3)`

(ii) `color(blue)(( z_1 - z_2)^3 = z_1^3 - 3z_1^2 z_2 + 3z_1z_2^2 - z_2^3)`

(iv) `color(blue)(z_1^2 – z_2^2 = (z_1 + z_2) *( z_1 – z_2))`

Q 3048801703

Express the following in the form of `a + bi:`

(i) `(-5i)(1/8 i)`

(ii) `(-i) (2i) (-1/8 i)^3`

(i) `(-5i)(1/8 i)`

(ii) `(-i) (2i) (-1/8 i)^3`

(i) `(-5i)(1/8 i) = (-5)/8 i^2 = (-5)/8 i^2 = (-5)/8 (-1) = 5/8+5/8 i 0`

(ii) `(-i) (2i) (-1/8 i)^3 = 2 xx 1/(8xx8xx8) xx i^5 = 1/(256) (i^2)^2 i = 1/256 i`

Q 3058801704

Express `(5 – 3i)^3` in the form `a + ib.`

We have, `(5 – 3i)^3 = 53 – 3 × 5^2 × (3i) + 3 × 5 (3i)^2 – (3i)^3`

`= 125 – 225i – 135 + 27i = – 10 – 198i.`

Q 3018001800

Express `(- sqrt3+sqrt(-2) ) (2 sqrt3-i)` in the form of `a + ib`

We have, `(- sqrt3 + sqrt(-2) ) (2 sqrt3 -i) = ( - sqrt3 + sqrt2 i)( 2 sqrt3-i)`

` = -6+sqrt3i+2 sqrt6i - sqrt2 i^2 = (-6+sqrt2) + sqrt3(1+2 sqrt2) i`

● Let `z = a + ib` be a complex number.

Then, `color(blue)("the modulus of z, denoted by")` `color(red)(| z |)`, is defined to be the non-negative real number `sqrt(a^2+B^2)` ,

i.e. `color(blue)(| z | = sqrt(a^2+b^2))`

● The conjugate of `z`, denoted as `color(blue)(barz)` , is the complex number `a – ib,` i.e., `barz = a – ib.`

For example `| 3+i | = sqrt(3^2+1^2) = sqrt(10) , | 2-5i| = sqrt(2^2+(-5)^2) = sqrt(29)` and `bar(3+i) = 3-i , bar(2-5i) = 2+5i , bar(-3i-5) = 3i-5`

● Observe that the multiplicative inverse of the non-zero complex number `z` is given by

`color(blue)(z^(-1) )= 1/(a+ib) = a/(a^2+b^2) +i (-b)/(a^2+b^2) = (a-ib)/(a^2+b^2) =color(blue)( barz/(|z|^2))` or `color(red)(z barz = | z|^2)`

● For any two complex numbers `z_1` and `z_2` , we have

(i) `color(blue)( | z_1 z_2 | = | z_1 | | z_2 |)`

(ii) `color(blue)(| z_1/z_2 | = (| z_1 |)/(| z_2 |))` provided ` | z_2 | ne 0`

(iii) `color(blue)(bar(z_1 z_2) = barz_1 barz_2)`

(iv) `color(blue)(bar(z_1 pm z_2) = barz_1 pm barz_2)`

(v) `color(blue)(bar ((z_1/z_2)) = barz_1/barz_2)` provided `z_2 ne 0`

Then, `color(blue)("the modulus of z, denoted by")` `color(red)(| z |)`, is defined to be the non-negative real number `sqrt(a^2+B^2)` ,

i.e. `color(blue)(| z | = sqrt(a^2+b^2))`

● The conjugate of `z`, denoted as `color(blue)(barz)` , is the complex number `a – ib,` i.e., `barz = a – ib.`

For example `| 3+i | = sqrt(3^2+1^2) = sqrt(10) , | 2-5i| = sqrt(2^2+(-5)^2) = sqrt(29)` and `bar(3+i) = 3-i , bar(2-5i) = 2+5i , bar(-3i-5) = 3i-5`

● Observe that the multiplicative inverse of the non-zero complex number `z` is given by

`color(blue)(z^(-1) )= 1/(a+ib) = a/(a^2+b^2) +i (-b)/(a^2+b^2) = (a-ib)/(a^2+b^2) =color(blue)( barz/(|z|^2))` or `color(red)(z barz = | z|^2)`

● For any two complex numbers `z_1` and `z_2` , we have

(i) `color(blue)( | z_1 z_2 | = | z_1 | | z_2 |)`

(ii) `color(blue)(| z_1/z_2 | = (| z_1 |)/(| z_2 |))` provided ` | z_2 | ne 0`

(iii) `color(blue)(bar(z_1 z_2) = barz_1 barz_2)`

(iv) `color(blue)(bar(z_1 pm z_2) = barz_1 pm barz_2)`

(v) `color(blue)(bar ((z_1/z_2)) = barz_1/barz_2)` provided `z_2 ne 0`

Q 3018112000

Find the multiplicative inverse of `2 – 3i`.

Let `z = 2 – 3i`

Then `z = 2 + 3i` and `z^2 = 2^2 + ( − 3)^2 =13`

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

`z^(-1) = barz/(| z|^2) = (2+3i)/(13) = 2/13+3/13 i`

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

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

` = (2+3i)/(2^2-(3i)^2) = (2+3i)/(13) = 2/13+3/13 i`

Q 3008112008

Express the following in the form `a + ib`

(i) `(5+sqrt2i)/(1- sqrt2 i)`

(ii) `i^(-35)`

(i) `(5+sqrt2i)/(1- sqrt2 i)`

(ii) `i^(-35)`

(i) We have, `(5+ sqrt2 i)/(1- sqrt2 i) = ( 5+ sqrt2 i)/(1-sqrt2i) xx ( 1+ sqrt2 i)/(1+ sqrt2 i)`

` = (5+5 sqrt2 i + sqrt2i -2)/(1- ( sqrt2i)^2)`

` = (3+6 sqrt2 i)/(1+2) = (3 (1+2sqrt2i))/3 = 1+2 sqrt2i`

(ii) `i^(-35) = 1/i^(35) = 1/{(i^2)^(17)i} = 1/(-i) xx i/i = i/(-i^2) = i`