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

#### Question

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

x^2-(a+bi)x+4+3i=0
(1)

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.

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
(2)

and

3-bx=0
(3)

Solving the second equation gives

x=\dfrac{3}{b}
(4)

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}
(5)

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}

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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