Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

let z = a + ib

you get (a+ib)^2 + 4(a-ib) + 4 = 0

expand

a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0

(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0

equate imaginary and real parts on both sides to get

(a^2 - b^2 + 4a + 4) = 0 ...1
and 2abi - 4bi = 0 => b = 0 or a = 2

solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = +/-4

so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so z = 2 + 4i or 2- 4i

solve for a 3x^2 +ax -(a^2 -1) = 0 so that all solutions to the equation are real and positive

comparing with the euation

Ax^2 + Bx + C ( as a is there in given equation)

we have

roots = (- B +/- sqrt(B^2-4AC))/ (2A)

and we take the - sign for the second one as if - sign gives positive then + sign being greater adding -B gives positive.

so -B - sqrt(B^2 - 4AC) > 0

or B < - sqrt(B^2-4AC)

put B = a , A = 3 and C = a^2-1


to get a < - sqrt(a^2 + 12(a^1-1)
< - sqrt(13a^2- 12)

but a is -ve and so negate both sides to get

- a > sqrt(13a^2-12)

or a^2 > 13a^2-12 or a^2 < 1 or -1 < a < 1 but as a is -ve -1 < a ..1

as B^2 >= 4AC so a^2 > - 12(a^2- 12) or 13a^2 > 12 or a^2 > (12/13) so a < - sqrt(12/13) ..2

so a = - sqrt(12/13) is the value

 

Prove that if a + b + c = 0 then (a^2 + b^2 + c^2)^2 = 2(a^4 + b^4 + c^4)

We have
(a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2a^2b^2 + 2 b^2 c^2 + 2 c^2 a^2 ... 1

Now a^2 b^2 + a^2 c^2 = a^2((b+c)^2 – 2bc) = a^4 – 2a^2bc .2 (as b+c = - a)

Similarly
b^2 c^2 + b^2 a^2 = b^4 – 2b^2ac ..3
c^2a^2 + c^2b^2 = c^4 – 2c^abc ...4

from (2) (3) and (4)
2(a^2b^2 + b^2 c^2 + c^2 a^2) = (a^4 + b^4 + c^4 – 2abc(a+b+c))
= a^4 + b^4 + c^4 ...5
as a+b+c = 0
Putting value of  2(a^2b^2 + b^2 c^2 + c^2 a^2) from (5) in (1) we get the result

show that √(1 + √(-3)) + √(1 - √(-3)) = √(6)

Let y = √(1 + √(-3)) + √(1 - √(-3))

so y^2 = (√(1 + √(-3)) + √(1 - √(-3)))^2
= (1 + √(-3)) + (1 - √(-3)) + 2 √((1 + √(-3))(1 - √(-3)))
= (1 + √(-3)) + (1 - √(-3)) + 2 √(1 - (-3)))
= 6

so y = √(6) as LHS is positive

hence √(1 + √(-3)) + √(1 - √(-3)) = √(6)

Tan [ cos^ -1 (1/√3) + sin^ -1(1/√3)]

we know cos ^- 1 (x) = pi/2 - sin ^-1 (x) ( you can find it from rt angle triangle)

or from cos t = sin (90-t) and taking arc cos

so cos ^- 1 (x) +- sin ^-1 (x) = pi/ 2

so cos^ -1 (1/√3) + sin^ -1(1/√3) = pi/ 2

and hence tan [ cos^ -1 (1/√3) + sin^ -1(1/√3)] = tan pi/2 = infinity
Note:
actually it is infinite(or to be more precise tends to be infinite) from left and - infinite for right so strictly speaking the value does not exist.

Prove these inequalities?

a) a^4 + b^4 >/= a^3b + ab^3

b) a^2 + b^2 + c^2 >/= ab + bc + ac


ans:
a)
we have following from AM GM enaquality

a^4 + (ab)^2 >= 2a^3.b .. 1
b^4 + (ab)^2 >= 2b^3.a ... 2
a^4 + b^4 >= 2(ab)^2 .. 3

add to get 2(a^4+b^4 ) + 2a^2b^2 >= 2a^3b + 2 ab^3 + 2 a^2 b^2
or (a^4+b^4 ) >= a^3b + ab^3

hence proved
b) we know a^2 + b^2 > = 2ab (from AM GM enaquality)
b^2 + c^2 >= 2bc
c^2 + a^2 >= 2ac

add to get 2(a^2 + b^2 + c^2) >= 2(ab+bc+ca)
or (a^2 + b^2 + c^2) >= ab + bc + ca

factorize:(x+1)(x+2)(x+3)(x+6)-3x^2

= (x+2)(x+3) ( x+ 1)(x+6) - 3x^2
= (x^2 + 5x + 6)(x^2 + 7x + 6) - 3x^2 ( we take the product such that constants in both products are same)

let x^2 + 6x + 6 = z

so we get ( z- x)(z + x) - 3x^2
= z^2 - x^2 - 3x^2
= z^2 - 4x^2
= (z+2x)(z-2x)
= (x^2 +8x + 6)(x^2+ 4x + 6)
cannot be factored further.

If vertices of triangle have integral coordinates, prove that it cannot be an equilateral triangle

Proof:

Without loss of generality we can chose the co-ordinates ( we can make this by proper shift of co-ordinates) as A= (0,0), B= (x,y), C=(a,b)

Now slope of AB = y/x is rational

Slope of AC = b/a is rational

So tan (BAC) = ((y/x) – (a/b))/ ( 1 + ay/(bx)) which is rationsl

As tan 60 = sqrt(3) there cannot be any point with rational coefficient so angle BAC cannot be 60 degree

As we cannot find any rational point on the kline at 60 degrees so getting an equilateral triangle is not possible

Hence proved

Show that 3^w+3^x+3^y+3^z is a perfect cube for infinitely many integers

we know

(a+1)^3 = a^3 + 3a^2 + 3a + 1

put a = 3^p to get

(3^p+1)^2 = 3^(3p) + 3^(2p+1) + 3^(p+1) + 1

the above is true for any p and by multiplying by (3^3)^q we get more results

one solution set is

w = 3p + 3q
x = 2p+1 + 3q
y = p + 1 + 3q
z = 3q   ( for p and q over integers)

Find the LCM of a^5+a^4+1 and a^5+a+1

First let us find the GCD

GCD ( a^5+ a^ 4+ 1, a^5 + a + 1)
= GCD( a^5 + a + 1, a^4- a) using GCD(a,b) = gcd(b , a -b)
= GCD(a^5+ a+ 1 , a^3(a-1))
= GCD(a^5+ a + 1, a- 1) as a is coprime to a^5 + a + 1
= 1 as a-1 is not a factor of a^5 + a + 1

so a^5+a^4+1 and a^5+a+1 are copimes

so LCM = product = (a^5+a^4+1)(a^5+ a+ 1)

factor the trinomial 2w^2 - w – 36

Ans:

This is of the form ax^2 + bx+ c with a != 0 and a!= 1
let

ax^2 + bx + c = (mx + n)(px + q) = mn x^2 + (mq+pn) x + nq
b is split into mq+pn and product = mqpn = ac

using that we multiply 2 and -36 to get -72

now 2 numbers to be chosen that product = 72 and sum = -1 (coefficient of w) they are 8 and - 9

so 2w^2 - w - 36 = 2w^2 + 8w - 9w - 36 = 2w(w+4) - 9(w+4) = (w+4)(2w-9)