because LHS is real so z + 2i is real and let z + 2i = x or z = x - 2i
so |x+1 - 2i |= x +2|
or $(x+1)^2 + 4 = (x+2)^2$
or $x^2 + 2x + 5 = x^2 + 4x + 4$ or $x = \frac{1}{2}$
$z = \frac{1}{2} - 2i$
Friday, February 3, 2017
Thursday, February 2, 2017
2017/004) If $x^2+px+1$ ia a factor of $ax^3 + bx +c $ then
a) $a^2+c^2= -ab$, b) $a^2-c^2 = - ab $, c) $a^2-c^2 = ab$, d) None of these
Solution
we have other factor linear and hence $dx + e$
so $ax^3+ bx + c = (x^2+px+1)(dx + e)$
comparing coefficient of $x^2$ we have d = a and constant term gives c =e
so $ax^3+ bx + c = (x^2+px+1)(ax + c) = ax^3 + x^2(c + ap) + x(a + cp) + x$
comparing coefficients of $x^2$ on both sided $c+ap=0\cdots(1)$
comparing coefficient of x both sides $ a+cp = b$ or $a^2 + acp = ab$ or $a^2-c^2 = ab$(using from (1)) Hence C
Solution
we have other factor linear and hence $dx + e$
so $ax^3+ bx + c = (x^2+px+1)(dx + e)$
comparing coefficient of $x^2$ we have d = a and constant term gives c =e
so $ax^3+ bx + c = (x^2+px+1)(ax + c) = ax^3 + x^2(c + ap) + x(a + cp) + x$
comparing coefficients of $x^2$ on both sided $c+ap=0\cdots(1)$
comparing coefficient of x both sides $ a+cp = b$ or $a^2 + acp = ab$ or $a^2-c^2 = ab$(using from (1)) Hence C
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