Complex

**Becks** · June 7th 2010, 05:34 AM

(a) Show that (3 – i)^2 = 8 – 6i.

(b) The quadratic equation
az^2 + bz + 10i = 0,
where a and b are real, has a root 3 – i.
(i) Show that a = 3 and find the value of b.
(ii) Determine the other root of the equation, giving your answer in the form p + iq.
(Hint for (ii): Use the fact that the sum of the roots in any quadratic is –b/a).

**bigli** · June 7th 2010, 05:51 AM

a) $(3-i)^2=3^2-2(3)(i)+i^2=9-6i-1=8-6i$ .

b)
i) $a(3-i)^2+b(3-i)+10i=0$ then $8a+3b=0 , 10-6a-b=0$ hence $a=3 , b=-8$

ii) $\frac{-b}{a}=(3-i)+(p+qi)=\frac{-(-8)}{3}$ then $p=\frac{-1}{3} , q=1$ hence, other root is $\frac{-1}{3}+i$ .

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