If $$z \in C$$, such that the quadratic equation $$x^2-zx+4+3i=0$$ has real root.
Find z with the least value of the modulus of the complex number.

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Uniteasy Admin · Accepted Answer

Let z= $$a+bi$$ when the equation has the real root. Then we have
$$x^2-(a+bi)x+4+3i=0$$n
in which $$a,b\in R$$. It can also be observed that $$ b
e 0$$, or the LHS will always be complex number when $$x\in R$$ and not equal to zero.
Expand bracket and rearrange the LHS by real part and imaginary part of the complex number.
$$x^2-ax+4+(3-bx)i=0$$
In order for the equation to hold true, both the real and imaginary parts have to equal to zero. Then the following two equations are established.
$$x^2-ax+4 = 0$$n
and
$$3-bx=0$$n
Solving the second equation gives
$$x=\dfrac{3}{b} $$n
Substituting the result of (4) to the first equation (2) gives
$$\Big( \dfrac{3}{b} \Big) ^2-a\Big( \dfrac{3}{b}\Big) +4 = 0$$
$$a=\dfrac{\Big( \dfrac{3}{b} \Big) ^2 +4}{\dfrac{3}{b}}=\dfrac{4b^2+9}{3b} $$n
Then we can express the value of the modulus in terms of $$a,b$$, and use the arithmetic inequality to find the minimum value.
$$|z|=a^2+b^2=\Big( \dfrac{4b^2+9}{3b} \Big) ^2+b^2$$
$$=\dfrac{25}{9}b^2 +\dfrac{9}{b^2}+8$$
$$\geq 2\sqrt{\dfrac{25}{9}b^2\cdot \dfrac{9}{b^2}} +8$$
$$=18$$
in which, if and only if $$\dfrac{25}{9}b^2 =\dfrac{9}{b^2}$$, that is $$b^2 = \dfrac{9}{5}$$, the iniquality becomes equality. Subsequently, the value of $$a^2$$ could be determined as.
$$a^2 = 18-\dfrac{9}{5} = \dfrac{81}{5} $$c
From the equation (5), it shows a and b always has the same sign. Therefore the complex number z with the least modulus is 
$$\pm\Big( \dfrac{9\sqrt{5} }{5}+\dfrac{3\sqrt{5} }{5}i\Big) $$c

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