106Q1. the mistake ur doing is like this
x2+1=0
=> x(x2+1) = 0
=> x=0
multiplying by zero wont be valid here, so ur getting an absurd answer.
Put z=0 in the original equation it doesnt satisfy
Q1. The equation Z3+ i Z -1 =0 has how many real roots ?
The answer given is 0.
BUT IF WE DO IT LIKE THIS :
multiplying both sides by Z' [ Z conjugate ]...
|Z|2 Z2 + i|Z|2 - Z' =0
then by inspection we get num of real roots as atleast one [ for Z=0 ]
so what exactly is the answer ?
Q2.If arg(Z1')=arg(Z2), then
[a] Z2 = k Z1-1 ( k>0)
(b) Z2 = k Z1 ( k>0)
[c] |Z2| = |Z'|
[d] None of these
Answer given :a
But by drawing its diagram , I get option b as answer...where am i wrong ?
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5 Answers
106
1ANS 1 :
Z3 + iZ - 1 = 0 ..... (1)
IF Z HAS REAL ROOTS THEN Z = Z' (Z BAR) FOR IMAGINARY PART TO BE 0.
SO TAKING CONJUGATE OF THE ENTIRE EXPRESSION :
(Z')3 + i' Z' - 1' = 0
AS Z = Z'
Z3 - i Z - 1 = 0 ..... (2)
SOLVING EQUATIONS (1) AND (2) :
Z3- 2 i = Z-1+1 = 1- 2 i
SO WE GET Z3 = 1 AND Z =0 WHICH IS HIGHLY CONTRADICTORY .... SO THIS EQUATION HAS NO REAL ROOTS.
1AND AS FAR AS THE 2ND QUESTION IS CONCERNED....THE ANSWER GIVEN IS TOTALLY WRONG.
IT SHOULD BE (B) AS YOU HAVE GOT ... NOT [a] .