Question

This problem features the art of finding complex square roots as well as solving a quadratic equation using a resulting formulation based on completing the quadratic square.

(a) Given that u²=-60+32i , express u in the form a+bi where a,b=R

(b) Hence, solve the equation z²-(3_2i)z+5-5i=0

Answer 1

(a) Let $u=a+bi.$ Then

$$u^2=(a+bi)^2=a^2-b^2+2abi$$

Given $u^2=-60+32i.$ Substitute

$$a^2-b^2+2abi=-60+32i$$

$$\begin{matrix} a^2-b^2=-60 \\ ab=16 \end{matrix}$$

$$\begin{matrix} a^2-\dfrac{256}{a^2}=-60 \\ b=\dfrac{16}{a} \end{matrix}$$

$$a^4+60a^2-256=0$$

$$D=(60)^2-4(1)(-256)=4624$$

$$a^2=\dfrac{-60\pm\sqrt{4624}}{2(1)}=-30\pm34$$

Since $a\in \R,$ we take $a^2=-30+34=4$

$$\begin{matrix} a=-2 \\ b=-8\end{matrix}$$

Or

$$\begin{matrix} a=2 \\ b=8\end{matrix}$$

$$u=-2-8i\ or\ u=2+8i$$

b)

$$z²-(3-2i)z+5-5i=0$$

$$D=(3-2i)^2-4(1)(5-5i)=9-4-12i-20+20i$$

$$=9-4-12i-20+20i=-15+8i$$

$$z=\dfrac{-(3-2i)\pm\sqrt{D}}{2(1)}$$

Let $u=a+bi.$ Then

$$u^2=(a+bi)^2=a^2-b^2+2abi$$

Suppose that $u^2=D=-15+8i.$ Substitute

$$a^2-b^2+2abi=-15+8i$$

$$\begin{matrix} a^2-b^2=-15 \\ ab=4 \end{matrix}$$

$$\begin{matrix} a^2-\dfrac{16}{a^2}=-15 \\ b=\dfrac{4}{a} \end{matrix}$$

$$a^4+15a^2-16=0$$

$$D=(15)^2-4(1)(-16)=289$$

$$a^2=\dfrac{-15\pm\sqrt{289}}{2(1)}=\dfrac{-15\pm17}{2}$$

Since $a\in \R,$ we take $a^2=\dfrac{-15+17}{2}=1$

$$\begin{matrix} a=-1 \\ b=-4\end{matrix}$$

Or

$$\begin{matrix} a=1 \\ b=4\end{matrix}$$

$$u=-1-4i\ or\ u=1+4i$$

$$z=\dfrac{-(3-2i)\pm\sqrt{(-1-4i)^2}}{2(1)}$$

$$=\dfrac{-3+2i\pm(-1-4i)}{2}$$

$$z_1=\dfrac{-3+2i+1+4i}{2}=-1+3i$$

$$z_2=\dfrac{-3+2i-1-4i}{2}=-2-i$$

$$z=\dfrac{-(3-2i)\pm\sqrt{(1+4i)^2}}{2(1)}$$

$$=\dfrac{-3+2i\pm(1+4i)}{2}$$

$$z_3=\dfrac{-3+2i-1-4i}{2}=-2-i$$

$$z_2=\dfrac{-3+2i+1+4i}{2}=-1+3i$$

$$z=-2-i\ or\ z=-1+3i$$