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complex number question (1 Viewer)

metalicarulez

New Member
Joined
Dec 9, 2003
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9
im havin trouble with these questions :

q1
3 - 2i is one root of x2 + bx + c = 0 where b and c are real
Find b and c

q2
x<sup>2</sup> + 6x + k = 0 has one root @ where Im(@) = 2
If k is real find both roots of the equation and the value of k

q3
1 - 2i is one root of x<sup>2</sup> - (3 + i)x + k = 0. Find k and the other root of the equation

thanks a lot!
 

spice girl

magic mirror
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Aug 10, 2002
Messages
785
all of these Q's use the rule that if the coefficients of a polynomial are real, then the complex solutions occur in conjugate pairs...
 

sammeh

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Oct 17, 2003
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so in q1 given that one root is 3 - 2i, the other must be 3 + 2i...?
which gives b = 6 and c = 13?

heh havent done complex polynomials yet, but no time like the present to learn!

continuing on, in q2 if one root has im(@) = 2, then the root @ = a + 2i, and the other root which i'll call & (degree 2 eqn, only 2 roots) = a - 2i.
solving for a by using @ + & = -b/1 = -6
@ + & = 2a = -6, a = -3
@ = -3 + 2i, & = -3 - 2i
and k = @& = (-3 + 2i)(-3 - 2i) = 13
so eqn is x<sup>2</sup> + 6x + 13 = 0 (which you would kno as soon as you find the roots, as they're the same as in q1...recycling questions, weak :p)

in q3 the roots wont occur in conjugate pairs, bcos the coefficiant of x<sup>1</sup> = (3 + i) which is not real, yah? which means its a simple "find the root" q, using same terminology as in q2;
@ = 1 - 2i
@ + & = 3 + i
& = 2 + 3i
k = @& = (1 - 2i)(2 + 3i) = 6 - i
eqn is x<sup>2</sup> - (3 + i)x + (6 - i) = 0

i sincerely doubt the validity of my answers, but if they're right, well i've learnt something today then :)
 
Last edited:

:: ck ::

Actuarial Boy
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Jan 1, 2003
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2004
all correct

last one is wrong i think

k = @& = (1 - 2i)(2 + 3i) = 6 - i
eqn is x2 - (3 + i)x + (6 - i) = 0
(1 - 2i)(2 + 3i) = (2 - (-6)) + i (3 - 4)
= 8 - i
not 6 - i :)

the rest u can do urself
 

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