Complex Part1

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Exercise 1

Express the following complex numbers in the form of a+bi\,

1) \left(3+4i\right)+\left(5-3i\right) 8+i\,

2) \left(2+7i\right)-\left(i-3\right) 5+6i\,

3) i\left(3+4i\right) -4+3i\,

  • \begin{align} i\left(3+4i\right)& =3i+4i^{2} \\ & = 3i-4 \\ & = 4-3i \\ \end{align}

4) \sqrt{2}i\left(\sqrt{2}-i\right) \sqrt{2}+2i\,

  • \begin{align} \sqrt{2}i\left(\sqrt{2}-i\right)& =2i-\sqrt{2}i^{2} \\ & =\sqrt{2}+2i\\ \end{align}

5) \left(3-i\right)\left(2+3i\right) 9+7i\,

  • \begin{align} \left(3-i\right)\left(2+3i\right)& =6+9i-2i-3i^{2} \\ & =9+7i \end{align}

6) \left(7-i\right)\left(4-3i\right) 25-25i\,

  • \begin{align} \left(7-i\right)\left(4-3i\right)& =28-4i-21i+3i^{2} \\ & =25-25i \end{align}

7) i\left(2+5i\right)\left(2-5i\right) 29i\,

  • \begin{align} i\left(2+5i\right)\left(2-5i\right)& =i\left(4-25i^{2}\right) \\ & =29i \end{align}

8) -2i\left(1-i\right)\left(2+i\right) -2-6i\,

  • \begin{align} -2i\left(1-i\right)\left(2+i\right)& =-2i\left(2-2i+i-i^{2}\right) \\ & =-2i\left(3-i\right)\\ & =-6i+2i^{2}\\ & =-2-6i \end{align}

9) \left(1+i\right)^{2} 2i\,

  • \begin{align} \left(1+i\right)^{2}& =1+2i+i^{2} \\ & =2i \end{align}

10) \left(2+7i\right)^{2} -45+28i\,

  • \begin{align} \left(2+7i\right)^{2}& =4+28i+49i^{2} \\ & =-45+28i \end{align}

11) \left(3-2i\right)^{2} 5-12i\,

  • \begin{align} \left(3-2i\right)^{2}& =9-12i+4i^{2} \\ & =5-12i \end{align}

12) \left(1-\sqrt{3}i\right)^{2} -2-2\sqrt{3}i\,

  • \begin{align} \left(1-\sqrt{3}i\right)^{2}& =1-2\sqrt{3}i+3i^{2} \\ & =-2-2\sqrt{3}i \end{align}

13) \left(1+i\right)^{3} -2+2i\,

  • \begin{align} \left(1+i\right)^{3}& =1+3i+3i^{2}+i^{3} \\ & =1+3i-3-i \\ & =-2+2i \end{align}

14) \left(2-3i\right)^{3} -46-9i\,

  • \begin{align} \left(2-3i\right)^{3}& =2^{3}+3\left(2^{2}\right)\left(-3i\right)+3\left(2\right)\left(-3i\right)^{2}+\left(-3i\right)^{3}\\ & =8-36i+54i^{2}-27i^{3}\\ & =8-36i-54+27i\\ & =-46-9i \end{align}

15) i\left(2-5i\right)^{2} 20-21i\,

  • \begin{align} i\left(2-5i\right)^{2}& =i\left(4-20i+25i^{2}\right)\\ & =i\left(-21-20i\right)\\ & =20-21i \end{align}

16) \left(1-i\right)\left(2-i\right)^{2} -1-7i\,

  • \begin{align} \left(1-i\right)\left(2-i\right)^{2}& =\left(1-i\right)\left(4-4i+i^{2}\right)\\ & =\left(1-i\right)\left(3-4i\right)\\ & =3-4i-3i+4i^{2}\\ & =-1-7i\\ \end{align}

17) \left(2+3i\right)^{2}\left(1-3i\right)^{2} 112-66i\,

  • \begin{align} \left(2+3i\right)^{2}\left(1-3i\right)^{2}& =\left(4+12i+9i^{2}\right)\left(1-6i+9i^{2}\right)\\ & =\left(-5+12i\right)\left(-8-6i\right)\\ & =40+30i-96i-72i^{2}\\ & =112-66i\\ \end{align}

18) \left(1-2i\right)^{3}\left(1+i\right)^{2} -4-22i\,

  • \begin{align} \left(1-2i\right)^{3}& =1+3\left(-2i\right)+3\left(-2i\right)^{2}+\left(-2i\right)^{3}\\ & =1-6i+12i^{2}-8i^{3}\\ & =1-6i-12+8i\\ & =-11+2i \\ \left(1+i\right)^{2}& =1+2i+i^{2}\\ & =2i \\ \therefore \left(1-2i\right)^{3}\left(1+i\right)^{2}& =\left(-11+2i\right)\left(2i\right)\\ & =-4-22i \end{align}

19) \frac{2}{i} -2i\,

  • \begin{align} \frac{2}{i}&=\frac{2}{i}\times\frac{i}{i}\\ &=\frac{2i}{-1}\\ &=-2i \end{align}

20) -\frac{3}{4i} \frac{3i}{4}\,

  • \begin{align} -\frac{3}{4i}&=-\frac{3}{4i}\times\frac{i}{i}\\ &=-\frac{3i}{\left(-4\right)}\\ &=\frac{3}{4}i \end{align}

21) \frac{2}{4+i} \frac{8}{17}-\frac{2}{17}i\,

  • \begin{align} \frac{2}{4+i}&=\frac{2}{4+i}\times\frac{4-i}{4-i}\\ &=\frac{8-2i}{16+1}\\ &=\frac{8}{17}-\frac{2}{17}i \end{align}

22) \frac{i}{i-1} \frac{1}{2}-\frac{1}{2}i\,

  • \begin{align} \frac{i}{i-1}&=\frac{-i}{1-i}\times\frac{1+i}{1+i}\\ &=\frac{-i+1}{1+1}\\ &=\frac{1}{2}-\frac{1}{2}i \end{align}

23) \frac{2+i}{3-2i} \frac{4}{13}+\frac{7}{13}i\,

  • \begin{align} \frac{2+i}{3-2i}&=\frac{2+i}{3-2i}\times\frac{3+2i}{3+2i}\\ &=\frac{6+7i+2i^{2}}{9+4}\\ &=\frac{4}{13}+\frac{7}{13}i \end{align}

24) \frac{4-5i}{2i+3} \frac{2}{13}-\frac{23}{13}i\,

  • \begin{align} \frac{4-5i}{2i+3}&=\frac{4-5i}{3+2i}\times\frac{3-2i}{3-2i}\\ &=\frac{12-23i+10i^{2}}{9+4}\\ &=\frac{2}{13}-\frac{23}{13}i \end{align}

25) -\frac{1}{\left(2+i\right)^{2}} -\frac{3}{25}+\frac{4}{25}i\,

  • \begin{align} -\frac{1}{\left(2+i\right)^{2}}&=-\frac{1}{4+4i+i^{2}}\\ &=-\frac{1}{3+4i}\times\frac{3-4i}{3-4i}\\ &=\frac{-3+4i}{9+16}\\ &=-\frac{3}{25}+\frac{4}{25}i \end{align}

26) \frac{1+i}{i\left(3+2i\right)^{2}} -\frac{7}{169}-\frac{17}{169}i\,

  • \begin{align} \frac{1+i}{i\left(3+2i\right)^{2}}&=\frac{1+i}{i\left(9+12i+4i^{2}\right)}\\ & =\frac{1+i}{i\left(5+12i\right)}\\ & =\frac{1+i}{-12+5i}\times\frac{-12-5i}{-12-5i}\\ & =\frac{-12-5i-12i-5i^{2}}{144+25}\\ &=-\frac{7}{169}-\frac{17}{169}i \end{align}

27) \frac{3}{1+i}\times\frac{2}{2-3i} \frac{15}{13}+\frac{3}{13}i

  • \begin{align} \frac{3}{1+i}\times\frac{2}{2-3i}& =\frac{6}{\left(1+i\right)\left(2-3i\right)}\\ & = \frac{6}{2-i-3i^{2}}\\ & = \frac{6}{5-i}\times \frac{5+i}{5+i}\\ & = \frac{30+6i}{25+1}\\ & = \frac{15}{13}+\frac{3}{13}i \end{align}

28) \frac{2-i}{2+i}\times\frac{1}{3-i} \frac{13}{50}-\frac{9}{50}i

  • \begin{align} \frac{2-i}{2+i}\times\frac{1}{3-i}& =\frac{2-i}{\left(2+i\right)\left(3-i\right)}\\ & = \frac{2-i}{6+i-i^{2}}\\ & = \frac{2-i}{7+i}\times \frac{7-i}{7-i}\\ & = \frac{14-9i+i^{2}}{49+1}\\ & = \frac{13}{50}-\frac{9}{50}i \end{align}

29) \frac{1}{2+i}+\frac{1}{3+2i} \frac{41}{65}-\frac{23}{65}i

  • \begin{align} \frac{1}{2+i}+\frac{1}{3+2i}& =\frac{3+2i+2+i}{\left(2+i\right)\left(3+2i\right)}\\ & = \frac{5+3i}{6+7i+2i^{2}}\\ & = \frac{5+3i}{4+7i}\times \frac{4-7i}{4-7i}\\ & = \frac{20-23i-21i^{2}}{16+49}\\ & = \frac{41}{65}-\frac{23}{65}i \end{align}

30) \frac{1-i}{1+i}+\frac{3i}{1+2i} \frac{6}{5}-\frac{2}{5}i

  • \begin{align} \frac{1-i}{1+i}+\frac{3i}{1+2i}& =\frac{\left(1-i\right)\left(1+2i\right)+3i\left(1+i\right)}{\left(1+i\right)\left(1+2i\right)}\\ & = \frac{1+i-2i^{2}+3i-3}{1+3i+2i^{2}}\\ & = \frac{4i}{-1+3i}\times\frac{-1-3i}{-1-3i}\\ & = \frac{-4i-12i^{2}}{1+9}\\ & = \frac{12-4i}{10}\\ & = \frac{6}{5}-\frac{2}{5}i \end{align}

Exercise 2

Given z=2-i\, and w=3+4i\,. Find, in the form of a+bi\,

1) z+w\, 5-3i\,

  • \begin{align} z+w &= 2-i + 3+4i \\ & =5-3i \end{align}

2) z-w\, -1-5i\,

  • \begin{align} z-w &= 2-i -\left(3+4i\right) \\ & =-1-5i \end{align}

3) zw\, 10+5i\,

  • \begin{align} zw &= \left(2-i\right)\left(3+4i\right) \\ & =6+5i-4i^{2} \\ & =10+5i \end{align}

4) \frac{z}{w}\, \frac{2}{25}-\frac{11}{25}i\,

  • \begin{align} \frac{z}{w} &= \frac{2-i}{3+4i}\times\frac{3-4i}{3-4i} \\ & = \frac{6-11i+4i^{2}}{9+16} \\ & =\frac{2}{25}-\frac{11}{25}i \end{align}

5) z^{\ast}-1\, 1+i\,

  • \begin{align} z^{\ast}-1 &= \left(2+i\right)-1 \\ & =1+i \end{align}

6) \left(z-1\right)^{\ast}\, 1+i\,

  • \begin{align} \left(z-1\right)^{\ast}&= \left(2-i-1\right)^{\ast} \\ &= \left(1-i\right)^{\ast}\\ & = 1+i \end{align}

7) \overline{zw}\, 10-5i\,

  • \begin{align} zw &= \left(2-i\right)\left(3+4i\right) \\ & =6+5i-4i^{2} \\ & =10+5i\\ \therefore \overline{zw}& = 10-5i \end{align}

8) \frac{z}{w^{\ast}}\, \frac{2}{5}+\frac{1}{5}i,

  • \begin{align} \frac{z}{w^{\ast}} &= \frac{2-i}{3-4i}\times\frac{3+4i}{3+4i} \\ & = \frac{6+5i-4i^{2}}{9+16} \\ & = \frac{10+5i}{25} \\ & =\frac{2}{5}+\frac{1}{5}i \end{align}

9) \left(z-1\right)\left(w+2\right)\, 9-i\,

  • \begin{align} \left(z-1\right)\left(w+2\right) &= \left(2-i-1\right)\left(3+4i+2\right) \\ &= \left(1-i\right)\left(5+4i\right) \\ &= 5-i-4i^{2} \\ & =9-i\\ \end{align}

10) \frac{z+i}{w-3}\, \frac{1}{2}i\,

  • \begin{align} \frac{z+i}{w-3} &= \frac{2-i+i}{3+4i-3} \\ &= \frac{2}{4i}\times\frac{i}{i} \\ &= \frac{i}{2i^{2}} \\ & =\frac{1}{2}i\\ \end{align}

11) \left(\frac{z^{2}}{iw}\right)^{\ast}\, -\frac{24}{25}-\frac{7}{25}i\,

  • \begin{align} \frac{z^{2}}{iw} &= \frac{\left(2-i\right)^{2}}{i\left(3+4i\right)} \\ &= \frac{4-4i+i^{2}}{3i+4i^{2}} \\ &= \frac{3-4i}{-4+3i}\times \frac{-4-3i}{-4-3i} \\ &= \frac{-12+7i+12i^{2}}{16+9}\\ &= -\frac{24}{25}+\frac{7}{25}i\\ \therefore \left(\frac{z^{2}}{iw}\right)^{\ast}&=-\frac{24}{25}-\frac{7}{25}i \end{align}

12) z^{2}-w^{2}\, 10-28i\,

  • \begin{align} z^{2}-w^{2} &= \left(2-i\right)^{2}-\left(3+4i\right)^{2} \\ &= \left(4-4i+i^{2}\right)-\left(9+24i+16i^{2}\right) \\ &= \left(3-4i\right)-\left(-7+24i\right) \\ &= 10-28i\\ \end{align}

13) \left(zw+i\right)^{2}\, 64+120i\,

  • \begin{align} \left(zw+i\right)^{2} &= \left[\left(2-i\right)\left(3+4i\right)+i\right]^{2} \\ &= \left(6+5i-4i^{2}+i\right)^{2} \\ &= \left(10+6i\right)^{2} \\ &= 100+120i+36i^{2} \\ & = 64+120i \end{align}

14) \frac{1}{\left(z-w\right)^{2}}\, -\frac{6}{169}-\frac{5}{338}i\,

  • \begin{align} \frac{1}{\left(z-w\right)^{2}} &= \frac{1}{\left[\left(2-i\right)-\left(3+4i\right)\right]^{2}} \\ &= \frac{1}{\left(-1-5i\right)^{2}} \\ &= \frac{1}{1+10i+25i^{2}} \\ &= \frac{1}{-24+10i}\times\frac{-24-10i}{-24-10i} \\ &=\frac{-24-10i}{576+100} \\ &=-\frac{6}{169}-\frac{5}{338}i \end{align}

15) \frac{1}{iz-1}-\frac{i}{2-w}\, \frac{4}{17}-\frac{15}{34}i\,

  • \begin{align} \frac{1}{iz-1}&= \frac{1}{i\left(2-i\right)-1}\\ &= \frac{1}{2i-i^{2}-1}\\ &= \frac{1}{2i}\times\frac{i}{i}\\ &= \frac{i}{2i^{2}}\\ &= -\frac{1}{2}i\\ \frac{i}{2-w} & =\frac{i}{2-\left(3+4i\right)}\\ & =\frac{i}{-1-4i}\times\frac{-1+4i}{-1+4i}\\ & = \frac{-i+4i^{2}}{1+16}\\ & = -\frac{4}{17}-\frac{1}{17}i\\ \therefore \frac{1}{iz-1}-\frac{i}{2-w} & = -\frac{1}{2}i-\left(-\frac{4}{17}-\frac{1}{17}i\right)\\ & =\frac{4}{17}-\frac{15}{34}i \end{align}

16) i\left(2z-3\right)^{2}\left(iw-3\right) -19+33i\,

  • \begin{align} i\left(2z-3\right)^{2}\left(iw-3\right) &= i\left[2\left(2-i\right)-3\right]^{2}\left[i\left(3+4i\right)-3\right] \\ &= i\left(1-2i\right)^{2}\left(-7+3i\right) \\ &= \left(1-4i+4i^{2}\right)\left(-7i+3i^{2}\right)\\ &= \left(-3-4i\right)\left(-3-7i\right) \\ &= 9+33i+28i^{2}\\ &=-19+33i \end{align}

17) \frac{1}{z^{2}w^{2}}+\left(z-3\right)^{2} -\frac{9997}{625}-\frac{4}{625}i\,

  • \begin{align} z^{2}w^{2}&=\left(2-i\right)^{2}\left(3+4i\right)^{2}\\ & =\left(4-4i+i^{2}\right)\left(9+24i+16i^{2}\right)\\ & =\left(3-4i\right)\left(-7+24i\right)\\ & = -21+100i-96i^{2} \\ & = 75+100i\\ \therefore \frac{1}{z^{2}w^{2}} & =\frac{1}{75+100i}\\ & =\frac{1}{25}\left(\frac{1}{3+4i}\right)\times\frac{3-4i}{3-4i}\\ &=\frac{1}{25}\left(\frac{3-4i}{9+16}\right)\\ &=\frac{3}{625}-\frac{4}{625}i\\ \left(z-3\right)^{2} & = \left(3+4i-3\right)^{2}\\ & = \left(4i\right)^{2}\\ & = -16 \\ \therefore \frac{1}{z^{2}w^{2}}+\left(z-3\right)^{2} & =\frac{3}{625}-\frac{4}{625}i-16\\ & =-\frac{9997}{625}-\frac{4}{625}i \end{align}
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