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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.2

(iii)

Solve:

(x+iy)2=2i−33+i(x+iy)^2=\frac{2i-3}{3+i}

Solution

First simplify the right-hand side:

2i−33+i=−3+2i3+i⋅3−i3−i=(−3+2i)(3−i)(3+i)(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)}{(3+i)(3-i)} \end{aligned} (−3+2i)(3−i)=−9+3i+6i−2i2=−7+9i(3+i)(3−i)=9+1=10\begin{aligned} (-3+2i)(3-i)&=-9+3i+6i-2i^2=-7+9i\\ (3+i)(3-i)&=9+1=10 \end{aligned}

So:

(x+iy)2=−710+910i(x+iy)^2=-\frac{7}{10}+\frac{9}{10}i

Equate parts:

x2−y2=−710(1)2xy=910⇒xy=920(2)\begin{aligned} & x^2-y^2=-\frac{7}{10} \quad (1)\\ & 2xy=\frac{9}{10} \Rightarrow xy=\frac{9}{20} \quad (2) \end{aligned}

Also,

(x2+y2)2=(x2−y2)2+(2xy)2=(710)2+(910)2=1310⇒ x2+y2=1310(3)\begin{aligned} (x^2+y^2)^2&=(x^2-y^2)^2+(2xy)^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}} \quad (3) \end{aligned}

Then:

x2=(x2+y2)+(x2−y2)2=1310−7102y2=(x2+y2)−(x2−y2)2=1310+7102\begin{aligned} x^2&=\frac{(x^2+y^2)+(x^2-y^2)}{2}=\frac{\sqrt{\frac{13}{10}}-\frac{7}{10}}{2}\\ y^2&=\frac{(x^2+y^2)-(x^2-y^2)}{2}=\frac{\sqrt{\frac{13}{10}}+\frac{7}{10}}{2} \end{aligned}

Since xy>0xy>0, xx and yy have the same sign.

x=±1310−7102,y=±1310+7102 (same sign)\boxed{x=\pm\sqrt{\frac{\sqrt{\frac{13}{10}}-\frac{7}{10}}{2}},\quad y=\pm\sqrt{\frac{\sqrt{\frac{13}{10}}+\frac{7}{10}}{2}}\ \text{(same sign)}}
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(ii)