Solving equation involving complex numbers
(a) Find real numbers $a$ and $b$ such that $(a+bi)^2 = -3-4i.$
(b) Hence solve the equation: $z^2+i\sqrt{3}z+i = 0$.
In the above question, I have solved part (a), with $a=\pm1$ and with $b=\mp2$, but I am not sure how to use this information to solve (b). I know they are relevant, because of the key word "hence," which dictates that I should use the information derived from part (a).
Please advise. Sorry in advance for any mistakes in labelling of the title and tags (I'm not very good at those, but am trying to improve on it)
$\endgroup$3 Answers
$\begingroup$By completing the square, we have:
$$z^2+i\sqrt{3}z+i = 0$$
$$z^2+i\sqrt{3}z-\dfrac{3}{4} = -\dfrac{3}{4}-i$$
$$\left(z+\dfrac{i\sqrt{3}}{2}\right)^2 = \dfrac{-3-4i}{4}$$
$$(2z+i\sqrt{3})^2 = -3-4i$$
Do you see how part (a) is relevant now?
$\endgroup$ 2 $\begingroup$However, I arrived at this:
$z^2+i\sqrt(3)z+i=0$
$z^2+\frac{i\sqrt(3)z}{2}-\frac34+i=0$
Taking $-\frac34+i$ to the other side results in:
$z^2+\frac{i\sqrt(3)z}{2}=\frac34-i$
Did I do something wrong?
$\endgroup$ 2 $\begingroup$Notice, $$(a+bi)^2=-3-4i$$ $$(a^2-b^2)+2iab=-3-4i$$ comparing real & imaginary parts, one should get $$a^2-b^2=-3\tag 1$$ $$2ab=-4\tag 2$$ solve (1) & (2) for the values of $a$ & $b$.
b) $$z^2+i\sqrt 3z+i=0$$ $$\left(z^2+2i\frac{\sqrt3}{2}z+\left(\frac{i\sqrt{3}}{2}\right)^2\right)-\left(\frac{i\sqrt{3}}{2}\right)^2+i=0$$ $$\left(z+\frac{i\sqrt{3}}{2}\right)^2=-\frac{3}{4}-i=\frac{-3-4i}{4}$$ $$\left(2z+i\sqrt 3\right)^2=-3-4i$$
$\endgroup$Alternative Method: use quadratic formula to find the roots of $z^2+i\sqrt 3z+i=0$ as follows $$z^2+i\sqrt 3z+i=0$$ $$z=\frac{-i\sqrt 3\pm\sqrt{(-i\sqrt 3)^2-4(1)(i)}}{2(1)}=\frac{-i\sqrt 3\pm i\sqrt{3+4i}}{2}$$
