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.

Collected in the board: Imaginary number

Steven Zheng posted 1 week ago


Let z= a+bi when the equation has the real root. Then we have


in which a,b\in R. It can also be observed that b\ne 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.


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



Solving the second equation gives


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}

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


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}

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)

Steven Zheng posted 1 week ago

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