First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.2

Solution

Use:

(x+iy)^2=x^2-y^2+2xy\,i

(i) $(x+iy)^2=25+60i$

\begin{aligned} x^2-y^2&=25 \\ 2xy&=60\Rightarrow xy=30 \end{aligned}

\begin{aligned} (x^2+y^2)^2&=(x^2-y^2)^2+(2xy)^2=25^2+60^2=4225 \\ \Rightarrow\ x^2+y^2&=65 \end{aligned}

\begin{aligned} x^2&=\frac{65+25}{2}=45\Rightarrow x=\pm 3\sqrt{5}\\ y^2&=\frac{65-25}{2}=20\Rightarrow y=\pm 2\sqrt{5} \end{aligned}

Since $xy>0$ , $(x,y)=(3\sqrt{5},2\sqrt{5})$ or $(-3\sqrt{5},-2\sqrt{5})$ .

(ii) $(x+iy)^2=64+48i$

\begin{aligned} x^2-y^2&=64 \\ 2xy&=48\Rightarrow xy=24 \end{aligned}

\begin{aligned} (x^2+y^2)^2&=64^2+48^2=6400 \\ \Rightarrow\ x^2+y^2&=80 \end{aligned}

\begin{aligned} x^2&=\frac{80+64}{2}=72\Rightarrow x=\pm 6\sqrt{2}\\ y^2&=\frac{80-64}{2}=8\Rightarrow y=\pm 2\sqrt{2} \end{aligned}

Since $xy>0$ , $(x,y)=(6\sqrt{2},2\sqrt{2})$ or $(-6\sqrt{2},-2\sqrt{2})$ .

(iii) $(x+iy)^2=\dfrac{2i-3}{3+i}$

\begin{aligned} \frac{2i-3}{3+i}&=\frac{-3+2i}{3+i}\cdot\frac{3-i}{3-i} =\frac{(-3+2i)(3-i)}{10} \\ (-3+2i)(3-i)&=-9+3i+6i-2i^2=-7+9i \end{aligned}

So $(x+iy)^2=-\dfrac{7}{10}+\dfrac{9}{10}i$ .

\begin{aligned} x^2-y^2&=-\frac{7}{10} \\ 2xy&=\frac{9}{10}\Rightarrow xy=\frac{9}{20} \end{aligned}

\begin{aligned} (x^2+y^2)^2&=\left(\frac{7}{10}\right)^2+\left(\frac{9}{10}\right)^2=\frac{13}{10} \\ \Rightarrow\ x^2+y^2&=\sqrt{\frac{13}{10}} \end{aligned}

\begin{aligned} x^2&=\frac{\sqrt{\frac{13}{10}}-\frac{7}{10}}{2}\\ y^2&=\frac{\sqrt{\frac{13}{10}}+\frac{7}{10}}{2} \end{aligned}

Since $xy>0$ , $x$ and $y$ have the same sign.

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