2015/089) lf $z= \frac{a+3i}{2+ai}$ , a is real.

Show that there is only one value of a for which arg z = $\frac{pi}{4}$  and find it
Solution
$arg(z) = \frac{pi}{4}$

z must be located at first quadrant...


now $\frac{a+3i}{2+ai} = \frac{(a+3i)(2-ai)}{(4+a^2)}$

= $\dfrac{5a + (6-a^2)i}{4+ a^2}$

Arg z = 1 when $5a = 6- a^2$
or $a^2 + 5a – 6 = 0$
$(a+6)(a-1) = 0$
a = -6 or 1
a = -6 means $arg(z) = \frac{3pi}{4}$ and hence this is invalid value and so $a = 1$ is the correct ans.
Refer to https://in.answers.yahoo.com/question/index?qid=20111204094456AAQhinr

2015/088)If a,b,c are real and distinct, then show that

$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\gt6abc$

taking 6abc to the left and reordering we get

$(a^2b^2-2bac + c^2) + (a^2c^2-2abc + b^2) + (b ^2c^2- 2abc + a^2)$

= $(ab-c)^2 + (ac-b)^2 + (bc-a)^2 \ge 0$

the expression can be zero only if $ab= c, ac = b , bc = a$ or $abc =$1 or $zero$

if a,b,c all are different then above cannot be zero

2015/087) If two vertices of an equilateral triangle are $(0,0),(3,\sqrt{3})$ then find the third vertex

one point is origin $(0,0)$ and another is $(3,\sqrt3)$

the length of the side = $\sqrt{12} = 2\sqrt3$

So it is at 30 degrees with x axis.

For it to be equilateral triangle the 2^nd side shall be at 60 degrees more or 60 degrees less
that is there are 2 triangle one is at 30 degrees more that is at 90 degrees which gives co-ordinates $(0, 2\sqrt3)$
another triangle at 30 degrees less that is at -30 degrees which gives co-ordinates $(3, -\sqrt{3})$

2015/086) Let $S=y^2+20y+12$ y is a positive integer.

What is the sum of all possible values of y for which S is a perfect square.

Solution

let $S = x^2$

so $x^2 = (y+10)^2 – 88$

or $88 = (y+10)^2-x^2 = (y+10 + x) (y+10-x)$

both the terms on RHS has to be even as one add and one even shall give fractional x and y

so $(y+10+x) = 44, (y + 10 – x) = 2$ giving $y = 13, x = 21$

or $(y+10+x) = 22, (y+10-x) = 4$ giving $y=3 , x = 9$

so sum of all possible values of $y = 16$

2015/085) Prove that

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)=2(x^3+y^3+z^3 – 3xyz)$
Solution 
knowing
$a^3+b^3 + c^3 – 3abc = \frac{1}{2}(a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)\cdots (1)$
putting $a = x + y , b= y + z, c= z+x$ we get

$(x+y)^3 + (y+z)^3 + (z+x)^3 -3(x+y)(y+z)(z+x)$

= $\frac{1}{2}(2x+2y+ 2z)((x-z)^2 + (y-x)^2 + (z-y)^2)$

= $2 * \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2 + (z-x)^2)$

= $2(x^3+y^3+z^3 – 3xyz)$ using (1)

2015/084) Prove that the locus of the center of the circle

$\frac{1}{2}(x^2 + y^2) + x \cos(\theta) + y \sin(\theta) - 4 = 0$ is $x^2 + y^2 = 1$

Solution

we have $(x^2 + 2 x \cos \theta) + (y^2 + 2y \sin \theta ) = 8$
or $(x^2 + 2 x \cos \theta+1) + (y^2 + 2y \sin \theta+ 1 ) = 10$

or $( x + \cos \theta)^2 + (y + \sin \theta ) ^2 = 10$
So the locus of the centre
$x = - \cos \theta$
$y = - \sin \theta$
to eliminate $\theta$
or $x^2 + y ^2 = 1$
above is the locus
so above is true