Complex Part2

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Exercise 3

Solve the following equations

1) $2z-i=3+zi\,$ $z=1+i\,$

$\begin{align} & 2z-i=3+zi \\ & 2z-zi=3+i \\ & z\left(2-i\right)=3+i \\ & z=\frac{3+i}{2-i}\times\frac{2+i}{2+i} \\ & \;=\frac{6+5i+i^{2}}{4+1} \\ & \;=\frac{5+5i}{5} \\ & \;=1+i \end{align}$

2) $\frac{iz}{1-i}=3+2i\,$ $z=-1-5i\,$

$\begin{align} & \frac{iz}{1-i}=3+2i \\ & iz=\left(3+2i\right)\left(1-i\right)\\ & \;=3-i-2i^{2}\\ & \;=5-i\\ & z=\frac{5-i}{i}\times\frac{i}{i}\\ & \; =\frac{5i-i^{2}}{-1}\\ & \; =-1-5i\\ \end{align}$

3) $\frac{1}{z-3}=2i\,$ $z=3-\frac{1}{2}i\,$

$\begin{align} & \frac{1}{z-3}=2i\\ & \frac{1}{2i}=z-3\\ & z= \frac{1}{2i}\times\frac{i}{i}+3\\ & \;=\frac{i}{-2}+3 \\ & \;=3-\frac{1}{2}i \\ \end{align}$

4) $\frac{3z+i}{i\left(z-2\right)}=2\,$ $z=\frac{10}{13}-\frac{15}{13}i\,$

$\begin{align} & \frac{3z+i}{i\left(z-2\right)}=2\\ & 3z+i=2i\left(z-2\right)\\ & 3z+i=2zi-4i \\ & 3z-2zi=-5i \\ & z\left(3-2i\right)=-5i\\ & z=\frac{-5i}{3-2i}\times\frac{3+2i}{3+2i}\\ & \;=\frac{-15i-10i^{2}}{9+4}\\ & \;=\frac{10}{13}-\frac{15}{13}i \end{align}$

5) $\left(2+3i\right)\left(a+bi\right)=i\left(1-3i\right)\,$ $a=\frac{9}{13}, b=-\frac{7}{13}\,$

$\begin{align} & \left(2+3i\right)\left(a+bi\right)=i\left(1-3i\right)\\ & a+bi=\frac{i\left(1-3i\right)}{2+3i}\\ & \qquad=\frac{i-3i^{2}}{2+3i}\\ & \qquad=\frac{3+i}{2+3i}\times\frac{2-3i}{2-3i}\\ & \qquad=\frac{6-7i-3i^{2}}{4+9}\\ & \qquad=\frac{9}{13}-\frac{7}{13}i\\ & \therefore a=\frac{9}{13}, b=-\frac{7}{13}\\ \end{align}$

6) $\left(a+bi\right)-\left(3-4i\right)=\frac{3+i}{2-5i}\,$ $a=\frac{88}{29}, b=-\frac{99}{29}\,$

$\begin{align} & \left(a+bi\right)-\left(3-4i\right)=\frac{3+i}{2-5i}\\ & a+bi=\frac{3+i}{2-5i}+3-4i\\ & \qquad=\frac{3+i}{2-5i}\times\frac{2+5i}{2+5i}+3-4i\\ & \qquad=\frac{6+17i+5i^{2}}{4+25}+3-4i\\ & \qquad=\frac{6+17i+5i^{2}}{4+25}+3-4i\\ & \qquad=\frac{1}{29}+\frac{17}{29}i+3-4i\\ & \qquad=\frac{88}{29}-\frac{99}{29}i\\ & \therefore a=\frac{88}{29}, b=-\frac{99}{29}\\ \end{align}$

7) $\left(3a+2i\right)+\left(5-bi\right)=10-3i\,$ $a=\frac{5}{3}, b=5\,$

$\begin{align} & \left(3a+2i\right)+\left(5-bi\right)=10-3i\\ & 3a+5+\left(2-b\right)i=10-3i\\ &\therefore 3a+5=10,\quad 2-b=-3 \\ & a=\frac{5}{3},\qquad \qquad b=5 \end{align}$

8) $\left(a+2i\right)\left(2+bi\right)=8+i\,$ $a=1, b=-3 \mbox{ or } a=3, b=-1\,$

$\begin{align} & \left(a+2i\right)\left(2+bi\right)=8+i\\ & 2a+abi+4i+2bi^{2}=8+i \\ & 2a-2b+\left(ab+4\right)i=8+i \\ &\therefore 2a-2b=8,\quad ab+4=1 \\ &\qquad a-b=4,\quad ab=-3\frac{\qquad}{}(2) \\ &\qquad a=4+b\frac{\qquad}{}(1) \\ &(1)\to(2):\\ & \left(4+b\right)b=-3 \\ & 4b+b^{2}=-3 \\ & b^{2}+4b+3=0 \\ & \left(b+3\right)\left(b+1\right)\\ & b=-3 \mbox { or } b=-1\\ &\mbox{When } b=-3 , a=1 \\ &\mbox{When } b=-1 , a=3 \\ &\therefore a=1, b=-3 \mbox{ or } a=3, b=-1 \end{align}$

9) $a\left(3-i\right)+b\left(1+i\right)=\frac{11-17i}{1-2i}\,$ $a=2, b=3\,$

$\begin{align} & a\left(3-i\right)+b\left(1+i\right)=\frac{11-17i}{1-2i}\\ & 3a-ai+b+bi=\frac{11-17i}{1-2i}\times\frac{1+2i}{1+2i}\\ & 3a+b+\left(-a+b\right)i =\frac{11+5i-34i^{2}}{1+4}\\ & \qquad\qquad\qquad\qquad =\frac{45+5i}{5}\\ & \qquad\qquad\qquad\qquad =9+i\\ &\therefore 3a+b=9\frac{\qquad}{}(1)\quad -a+b=1\frac{\qquad}{}(2) \\ & (1)-(2):\\ & 4a=8, \therefore a=2, b=3\\ \end{align}$

10) $\frac{a-i}{4+bi}=1-i\,$ $a=7, b=3\,$

$\begin{align} & \frac{a-i}{4+bi}=1-i\\ & a-i=\left(1-i\right)\left(4+bi\right)\\ & a-i=4+bi-4i-bi^{2}\\ & a-i=4+b+\left(b-4\right)i\\ &\therefore a=4+b,\quad -1=b-4 \\ & \qquad \qquad \qquad \quad b=3\\ & \therefore a=7\\ &\therefore a=7, b=3\\ \end{align}$

11) $w+z=5-i,w-2z=\frac{-3+14i}{2-i}\,$ $w= 2+i, z=3-2i\,$

$\begin{align} & w+z=5-i\frac{\qquad}{}(1)\qquad w-2z=\frac{-3+14i}{2-i}\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad =\frac{-3+14i}{2-i}\times\frac{2+i}{2+i} \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad =\frac{-6+25i+14i^{2}}{4+1} \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad =\frac{-20+25i}{5} \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad =-4+5i\frac{\qquad}{}(2) \\ &(1)-(2):\\ & 3z=9-6i \\ & z=3-2i\frac{\qquad}{}(3) \\ &(3)\to(1):\\ & w+3-2i=5-i \\ & w= 2+i\\ &\therefore w= 2+i, z=3-2i \\ \end{align}$

12) $iz-2w=3,z-\left(2+i\right)w=12-11i\,$ $w= 4i, z=8-3i\,$

$\begin{align} & iz-2w=3\frac{\qquad}{}(1)\qquad z-\left(2+i\right)w=12-11i\\ & \qquad\qquad\qquad\qquad\qquad\quad iz-\left(2i+i^{2}\right)w=12i-11i^{2}\\ & \qquad\qquad\qquad\qquad\qquad\quad iz-\left(-1+2i\right)w=11+12i\\ & \qquad\qquad\qquad\qquad\qquad\quad iz+\left(1-2i\right)w=11+12i\frac{\qquad}{}(2)\\ & (2)-(1):\\ & \left(1-2i\right)w+2w=11+12i-3\\ & \left(3-2i\right)w=8+12i\\ & w=\frac{8+12i}{3-2i}\times\frac{3+2i}{3+2i}\\ & \;=\frac{24+52i+24i^{2}}{9+4}\\ & \;=4i\frac{\qquad}{}(3)\\ & (3)\to(1):\\ & iz-8i=3\\ & z=\frac{3+8i}{i}\times\frac{i}{i}\\ & \;=\frac{3i+8i^{2}}{-1}\\ & \;=8-3i\\ &\therefore w= 4i, z=8-3i \\ \end{align}$

Exercise 4

1) Find $\sqrt{-3+4i}$ $1+2i \mbox { or } -1-2i\,$

$\begin{align} & \mbox{Let }\sqrt{-3+4i}=a+bi\\ & \quad \therefore -3+4i=\left(a+bi\right)^{2}\\ & \qquad \qquad \qquad =a^{2}+2abi+b^{2}i^{2}\\ & \qquad \qquad \qquad =a^{2}-b^{2}+2abi\\ & \therefore a^{2}-b^{2}=-3\frac{\qquad}{}(1) \qquad 2ab=4\\ & \qquad \qquad \qquad \qquad \qquad \qquad b=\frac{2}{a}\frac{\qquad}{}(2) \\ & (2)\to(1):\\ & a^{2}-\left(\frac{2}{a}\right)^{2}=-3\\ & a^{2}-\frac{4}{a^{2}}=-3\\ & a^{4}-4=-3a^{2}\\ & a^{4}+3a^{2}-4=0\\ & \left(a^{2}-1\right)\left(a^{2}+4\right)=0\\ & a^{2}=1 \mbox{ or } a^{2}=-4 \mbox{ (rejected) }\\ & \therefore a=\pm 1 \\ & \mbox{When } a=1, b=2 \\ & \mbox{When } a=-1, b=-2 \\ & \therefore \sqrt{-3+4i} =1+2i \mbox { or } -1-2i \end{align}$

2) Express $\sqrt{5+12i}$ in the form of $a+bi\,$ $3+2i \mbox { or } -3-2i\,$

$\begin{align} & \mbox{Let }\sqrt{5+12i}=a+bi\\ & \quad \therefore 5+12i=\left(a+bi\right)^{2}\\ & \qquad \qquad \qquad =a^{2}+2abi+b^{2}i^{2}\\ & \qquad \qquad \qquad =a^{2}-b^{2}+2abi\\ & \therefore a^{2}-b^{2}=5\frac{\qquad}{}(1) \qquad 2ab=12\\ & \qquad \qquad \qquad \qquad \qquad \qquad b=\frac{6}{a}\frac{\qquad}{}(2) \\ & (2)\to(1):\\ & a^{2}-\left(\frac{6}{a}\right)^{2}=5\\ & a^{2}-\frac{36}{a^{2}}=5\\ & a^{4}-36=5a^{2}\\ & a^{4}-5a^{2}-36=0\\ & \left(a^{2}-9\right)\left(a^{2}+4\right)=0\\ & a^{2}=9 \mbox{ or } a^{2}=-4 \mbox{ (rejected) }\\ & \therefore a=\pm 3 \\ & \mbox{When } a=3, b=2 \\ & \mbox{When } a=-3, b=-2 \\ & \therefore \sqrt{5+12i} =3+2i \mbox { or } -3-2i \end{align}$

3) Given $\left(a+bi\right)^{2}=-2i$ , where $a\,$ and $b\,$ are real, find possible values of $a\,$ and $b\,$ $a=1, b=-1 \mbox{ or } a=-1, b=1\,$

$\begin{align} & \left(a+bi\right)^{2}=-2i\\ & a^{2}+2abi+b^{2}i^{2}=-2i\\ & a^{2}-b^{2}+2abi=-2i\\ & \therefore a^{2}-b^{2}=0\frac{\qquad}{}(1) \qquad 2ab=-2\\ & \qquad \qquad \qquad \qquad \qquad \qquad b=-\frac{1}{a}\frac{\qquad}{}(2) \\ & (2)\to(1):\\ & a^{2}-\left(-\frac{1}{a}\right)^{2}=0\\ & a^{2}-\frac{1}{a^{2}}=0\\ & a^{2}=\frac{1}{a^{2}}\\ & a^{4}=1\\ & a^{2}=1 \mbox{ or } a^{2}=-1 \mbox{ (rejected) }\\ & \therefore a=\pm 1 \\ & \mbox{When } a=1, b=-1 \\ & \mbox{When } a=-1, b=1 \\ & a=1, b=-1 \mbox{ or } a=-1, b=1 \\ \end{align}$

4) Find $z\,$ if $z^{2}=1+2\sqrt{2}i$ $z=\sqrt{2}+i \mbox { or } -\sqrt{2}-i\,$

$\begin{align} & \mbox{Let }z=a+bi\\ & z^{2}=1+2\sqrt{2}i \\ & \therefore \left(a+bi\right)^{2}=1+2\sqrt{2}i\\ & a^{2}+2abi+b^{2}i^{2}=1+2\sqrt{2}i\\ & a^{2}-b^{2}+2abi=1+2\sqrt{2}i\\ & \therefore a^{2}-b^{2}=1\frac{\qquad}{}(1) \qquad 2ab=2\sqrt{2}\\ & \qquad \qquad \qquad \qquad \qquad \qquad b=\frac{\sqrt{2}}{a}\frac{\qquad}{}(2) \\ & (2)\to(1):\\ & a^{2}-\left(\frac{\sqrt{2}}{a}\right)^{2}=1\\ & a^{2}-\frac{2}{a^{2}}=1\\ & a^{4}-2=a^{2}\\ & a^{4}-a^{2}-2=0\\ & \left(a^{2}-2\right)\left(a^{2}+1\right)=0\\ & a^{2}=2 \mbox{ or } a^{2}=-1 \mbox{ (rejected) }\\ & \therefore a=\pm \sqrt{2} \\ & \mbox{When } a=\sqrt{2}, b=1 \\ & \mbox{When } a=-\sqrt{2}, b=-1 \\ & \therefore z =\sqrt{2}+i \mbox { or } -\sqrt{2}-i \end{align}$

Exercise 5

Find all complex number $z\,$ such that

1) $z^{2}=1\,$ $z= 1 \mbox { or } -1\,$

$\begin{align} & \mbox{Let } z=x+yi \\ & \left(x+yi\right)^{2}=1 \\ & x^{2}+2xyi+y^{2}i^{2}=1 \\ & x^{2}-y^{2}+2xyi=1\\ & \therefore x^{2}-y^{2}=1,\quad 2xy=0 \\ & \qquad \qquad \qquad \qquad x=0 \mbox{ or } y=0\\ & \mbox{When } x=0\\ & 0^{2}-y^{2}=1 \\ & \therefore y^{2}=-1 \mbox{(rejected)} \\ & \mbox{When } y=0\\ & x^{2}=1 \\ & \therefore x=\pm 1 \\ & \therefore z= 1 \mbox { or } -1 \end{align}$

2) $z^{2}=i\,$ $z=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i \mbox{ or } -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$

$\begin{align} & \mbox{Let } z=x+yi \\ & \left(x+yi\right)^{2}=i \\ & x^{2}+2xyi+y^{2}i^{2}=i \\ & x^{2}-y^{2}+2xyi=i\\ & \therefore x^{2}-y^{2}=0\frac{\qquad}{}(1) \quad 2xy=1 \\ & \qquad \qquad \qquad \qquad \qquad \quad y=\frac{1}{2x}\frac{\qquad}{}(2)\\ & (2)\to(1) \\ & x^{2}-\left(\frac{1}{2x}\right)^{2}=0 \\ & x^{2}-\frac{1}{4x^{2}}=0 \\ & x^{2}=\frac{1}{4x^{2}} \\ & x^{4}=\frac{1}{4} \\ & \therefore x^{2} =\pm \frac{1}{2} \\ & \therefore x^{2} = \frac{1}{2} \mbox{ or } -\frac{1}{2}\mbox{(rejected)} \\ & \therefore x =\pm \sqrt{\frac{1}{2}} \\ & \mbox{When } x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\\ & y=\frac{1}{2\sqrt{\frac{1}{2}}}=\frac{\sqrt{2}}{2} \\ & \mbox{When } x=-\sqrt{\frac{1}{2}}=-\frac{\sqrt{2}}{2}\\ & y=\frac{1}{2\left(-\sqrt{\frac{1}{2}}\right)}=-\frac{\sqrt{2}}{2} \\ & \therefore z=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i \mbox{ or } -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i \\ \end{align}$

3) $z^{2}=z\,$ $z= 1 \mbox { or } 0\,$

$\begin{align} & \mbox{Let } z=x+yi \\ & \left(x+yi\right)^{2}=x+yi \\ & x^{2}+2xyi+y^{2}i^{2}=x+yi \\ & x^{2}-y^{2}+2xyi=x+yi\\ & \therefore x^{2}-y^{2}=x,\quad 2xy=y \\ & \qquad \qquad \qquad \qquad 2xy-y=0\\ & \qquad \qquad \qquad \qquad y\left(2x-1\right)=0\\ & \qquad \qquad \qquad \qquad y=0 \mbox{ or } x=\frac{1}{2}\\ & \mbox{When } y=0\\ & x^{2}=x \\ & x^{2}-x=0 \\ & x\left(x-1\right)=0 \\ & \therefore x=1 \mbox{ or } x=0\\ & \mbox{When } x=\frac{1}{2}\\ & \frac{1}{4}-y^{2}=\frac{1}{2} \\ & \therefore y^{2}=-\frac{1}{2}\mbox { (rejected) } \\ & \therefore z= 1 \mbox { or } 0 \end{align}$

4) $z^{3}=1\,$ $z= 1, -\frac{1}{2}+\frac{\sqrt{3}}{2}i \mbox { or } -\frac{1}{2}-\frac{\sqrt{3}}{2}i\,$

$\begin{align} & \mbox{Let } z=x+yi \\ & \left(x+yi\right)^{3}=1 \\ & x^{3}+3x^{2}\left(yi\right)+3x\left(yi\right)^{2}+\left(yi\right)^{3}=1 \\ & x^{3}+3x^{2}yi+3xy^{2}i^{2}+y^{3}i^{3}=1 \\ & x^{3}+3x^{2}yi-3xy^{2}-y^{3}i=1 \\ & x^{3}-3xy^{2}+\left(3x^{2}y-y^{3}\right)i=1 \\ & \therefore x^{3}-3xy^{2}=1, \quad 3x^{2}y-y^{3}=0 \\ & \qquad \qquad \qquad \qquad \quad y\left(3x^{2}-y^{2}\right)=0\\ & \qquad \qquad \qquad \qquad \quad y=0 \mbox{ or } 3x^{2}=y^{2}\\ & \mbox{When } y=0\\ & x^{3}=1 \\ & x=1 \\ & \mbox{When } 3x^{2}=y^{2}\\ & x^{3}-3x\left(3x^{2}\right)=1 \\ & -8x^{3}=1 \\ & x^{3}=-\frac{1}{8} \\ & x=-\frac{1}{2} \\ & \therefore y^{2}=\frac{3}{4}\\ & y=\pm\frac{\sqrt{3}}{2}\\ & \therefore z= 1, -\frac{1}{2}+\frac{\sqrt{3}}{2}i \mbox { or } -\frac{1}{2}-\frac{\sqrt{3}}{2}i \end{align}$

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