IIT JEE , AIEEE Forum - Mathematics , Algebra - AWESOME QUESTIONS (not solved yet) IN COMPLEX NO.

shubham_sachdeva

first of all questions i m giving here are word to word .. so qns. are not wrong..

1. Find minimum |z| satisfying mod(z + 1/z)

2. find the condition on the complex constants

and

if z^2 +

z +

= 0 has a real root.

3. find the range of values of

belongs to R for which the eqn z +

| z-1| + 2i = 0 , z = x +iy has a soln. also find the soln.

4. prove that mod(z/|z| -1) =< |arg(z)|

5.if (x^2 + a^2)(x^2 + b^2)(x^2 + c^2) = {f(x)}^2 + {g(x)}^2

find f(x) & g(x) where f(x) is a polynomial of the third degree....

hope u will enjoy these questions...

shubham_sachdeva

noone here to take the challenge...
arrey they r simple seriouslyy....

joyfrancis

in Q2) i'm just getting that both alpha and beta should be purely real

so maybe the ans is

,

R

shubham_sachdeva

nopes the answer is very very long.......

vinu

Hi shubham,

2)if p were tht real root...then p²+

p+

= 0 ;

so, Im(

) / Im(

) = -p ; p²+p*Re(

) = -Re(

) ;

substitute value of p frm 1st relation into the 2nd relation....ans.

1) if u meant lz+1/zl = p(say)....

then check out the ans....

u know lz+1/zl <= lzl+1/lzl ....{let lzl be x}

so,x²-px+1 >= 0

therefore min value of x is ....u find it out

vinu

4) Put z = lzlcis

;

so LHS = lcis

-1l = 2*l sin

/2 l * l i*cis

/2 l..........{also, l i*cis

/2 l = 1 } ;

= 2*lsin

/2l ;

<= l

l = l arg(z) l.............ans.

3) given : x+i(y+2)+

*

[(x-1)²+y²] = 0........

also,given tht

is a real no...

so, y = -2 ; x+

*

[(x-1)²+y²] = 0 ;

i.e; x²(

²-1) - 2x

²+ 5 = 0 ;

x is real...so, discriminant >=0.........

⁴-5

²+ 5 >= 0 ...u can find range of

frm this.......ans.

pottermania1990

hi bro,

1)ans is |Z|=1

shubham_sachdeva

oye ye galat baat hain bhai .......answers pura nikal ke do yaar vinu...jo homework tumne diya hain na mujhe vaha taq maine bhi ker liya hain....ye homework bada mushkil hain.......:)

and pottermania answer is not 1

catch_arnnie

i'll rewrite the solution again...

but i'm getting the range as [-sqrt(5)/2 , sqrt(5)/2 ]

???

shubham_sachdeva

nopes ....answer is [root 5/2 , 1)

catch_arnnie

plz see my method & tell me what's wrong with it...

shubham_sachdeva

u r wrong where u have evaluated the modulus u forget to take the square root :)

catch_arnnie

oh damn!!
thanks!!

fused_bulb

ur 1st ques is incomplete...

lemme complete it for u...

probably...u wanna ask find min. mod(z) satisfying mod( z + 1/z ) =a

where a is some constant..

here is the ans.

mod ( z + 1/z ) = a

implies mod [ 1/z - ( -z ) ] = a

implies [ 1 / mod(z) ] - mod( -z ) < = a.......( triangle inequality )

let mod(z) = r.
also..mod ( -z ) = mod ( z ) = r..

therefore

(1/r) - r <=a

the equality wud be satisfied for r(minimum)

1/r(min) - r(min) = a

solve this quadratic to get r(min) which is nothin but z(min)

u will get z ( min ) in terms of a ( the constant used )..

just put the value of a to get the answer...

and dont ask me to solve the quadratic..

every child knows how to solve a quadratic...

and do rate me...

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