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.