Consider a quadratic equation $az^2+bz+c=0$, where $a, b, c$ are complex number.
Then condition that above equation has one purely imaginary root
(A)$(a\bar b+\bar ab)(b\bar c+ \bar b c )+(c \bar a -\bar c a)^2=0$
(B)$(a\bar b-\bar ab)(b\bar c+ \bar b c )+(c \bar a -\bar c a)^2=0$
(c)$(a\bar b-\bar ab)(b\bar c- \bar b c )+(c \bar a -\bar c a)^2=0$
(D) None of these
My solution
Note: $i=\sqrt{-1}$ and $k$ is real number.
$a=a_{1}+i\cdot a_{2},\;b=b_{1}+i\cdot b_{2},\;c=c_{1}+i\cdot c_{2}$
Let purely imaginary root $z=k\cdot i\;$ since $z$ is root of above equation it must satisfy the equation so equation turns into $(a_{1}+i\cdot a_{2})(ki)^2+(b_{1}+i\cdot b_{2})(ki)+(c_{1}+i\cdot c_{2})=0$
Comparing real and imaginary part of left hand side with real and imaginary part of right hand side. I obtained the equations
$$\begin{align}
a_{1}k^2+b_{2}k-c_{1}&=0\qquad(1)\\
\text{and }\quad a_2k^2-b_1k-c_{2}&=0 \qquad(2)
\end{align}
$$
Equations $(1)$ and $(2)$ must have both roots common because we want only one value of $k$.
After applying condition for both roots common I obtained $\dfrac{a{_1}}{a_{2}}=\dfrac{-b_2}{b_1}=\dfrac{c_1}{c_2} \qquad(3)$
Using $(3)$ I obtained $a_1b_1+a_2b_2=b_1c_1+b_2c_2=a_2c_1-a_1c_2=0$
Now using $(a\bar b+\bar a b)=2\Re(a\bar b)=2(a_1b_1+a_2b_2),\;$ similary $(b\bar c+ \bar b c )=2\Re(b\bar c)=2(b_1c_1+b_2c_2),\;(c \bar a -\bar c a)=2i\;(\Im (c \bar a))=2i\;(a_2c_1-a_1c_2)$.
Now $0\cdot 0=0^2\implies (a_1b_1+a_2b_2)(b_1c_1+b_2c_2)=(a_2c_1-a_1c_2)^2$
$\implies \dfrac{(a\bar b+\bar a b)}{2}\dfrac{(b\bar c+ \bar b c )}{2}=\bigg(\dfrac{c \bar a -\bar c a}{2i}\bigg)^2$
Hence $$(a\bar b+\bar ab)(b\bar c+ \bar b c )+(c \bar a -\bar c a)^2=0$$
Can anyone Verify me If I made mistake anywhere or I did some mathematical blunder
I know one other method to solve this problem. I just want to know If there anything mathematically wrong in my solution. Because My teacher said this problem cannot be solved using $z=ki$. He said we must take conjugate of original equation.
Related Question
Consider a quadratic equation $az^2+bz+c=0$ where a,b,c are complex numbers. Prove that the equation has one purely imaginary root is given …
Best Answer
This answer shows that $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$ is a necessary condition.
$iz^2-i=i(z+1)(z-1)=0$ shows that it is not a sufficient condition.
From the following part :
it seems that you think "equation has one purely imaginary root" means that "equation has a purely imaginary double root".
I think that you have to deal with the case $a_2b_1c_2=0$ separately.
Another way is to start with $a(z-ki)(z-ki)=0$. Since $b=-2kia$ and $c=-ak^2$, we have $(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$.
Using your idea, we can show that if $az^2+bz+c=0$ has at least one purely imaginary root, then $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$.
You have $$a_1k^2+b_2k-c_1=0\tag1$$ $$a_2k^2-b_1k-c_2=0\tag2$$
Now, $(1)\times a_2-(2)\times a_1$ gives $$(a_2b_2+a_1b_1)k=a_2c_1-a_1c_2$$ which can be written as $$\frac{a\bar b+\bar a b}{2}k=-\frac{c \bar a -\bar c a}{2i}$$ i.e. $$i(a\bar b+\bar a b)k=-(c \bar a -\bar c a)\tag3$$
Also, $(1)\times c_2-(2)\times c_1$ gives
$$(a_2c_1-a_1c_2)k^2=(b_1c_1+b_2c_2)k\tag4$$
If $k=0$, then we have $c=0$ for which $(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$ holds.
If $k\not=0$, then dividing the both sides of $(4)$ by $k$ gives $$(a_2c_1-a_1c_c)k=b_1c_1+b_2c_2$$ which can be written as $$-\frac{c \bar a -\bar c a}{2i}k=\frac{b\bar c+ \bar b c}{2}$$ i.e. $$(c \bar a -\bar c a)k=-i(b\bar c+ \bar b c)\tag5$$
Multiplying the both sides of $(3)$ by $(c \bar a -\bar c a)$ gives $$i(a\bar b+\bar a b)\color{red}{(c \bar a -\bar c a)k}=-(c \bar a -\bar c a)^2$$
Finally, using $(5)$ to eliminate $k$, we get $$i(a\bar b+\bar a b)\color{red}{\bigg(-i(b\bar c+ \bar b c)\bigg)}=-(c \bar a -\bar c a)^2$$ i.e. $$(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$$