Express the following complex numbers in the standard form a + i b:
Question:

Express the following complex numbers in the standard form a + i b:

(i) $(1+i)(1+2 i)$

(ii) $\frac{3+2 i}{-2+i}$

 

(iii) $\frac{1}{(2+i)^{2}}$

(iv) $\frac{1-i}{1+i}$

 

(v) $\frac{(2+i)^{3}}{2+3 i}$

(vi) $\frac{(1+i)(1+\sqrt{3} i)}{1-i}$

(vii) $\frac{2+3 i}{4+5 i}$

 

(viii) $\frac{(1-i)^{3}}{1-i^{3}}$

(ix) $(1+2 i)^{-3}$

 

(x) $\frac{3-4 i}{(4-2 i)(1+i)}$

(xi) $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{1-4 i}{5+i}\right)$

(xii) $\frac{5+\sqrt{2} i}{1-2 \sqrt{i}}$(i) $(1+i)(1+2 i)$

Solution:

(i) $(1+i)(1+2 i)$

$=1+2 i+i+2 i^{2}$

$=1+3 i-2 \quad\left(\because i^{2}=-1\right)$

$=-1+3 i$

(ii) $\frac{3+2 i}{-2+i}$

$=\frac{3+2 i}{-2+i} \times \frac{-2-i}{-2-i}$

$=\frac{-6-3 i-4 i-2 i^{2}}{4-i^{2}} \quad\left(\because i^{2}=-1\right)$

$=\frac{-6-7 i+2}{4+1}$

$=\frac{-4-7 i}{5}$

$=\frac{-4}{5}-\frac{7}{5} i$

(iii) $\frac{1}{(2+i)^{2}}$

$=\frac{1}{4+i^{2}+4 i} \quad\left(\because i^{2}=-1\right)$

$=\frac{1}{3+4 i}$

$=\frac{1}{3+4 i} \times \frac{3-4 i}{3-4 i}$

$=\frac{3-4 i}{9-16 i^{2}}$

$=\frac{3-4 i}{9+16}$

$=\frac{3}{25}-\frac{4}{25} i$

(iv) $\frac{1-i}{1+i}$

$=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$=\frac{1+i^{2}-2 i}{1-i^{2}} \quad\left(\because i^{2}=-1\right)$

$=\frac{-2 i}{2}$

$=-i$

(v) $\frac{(2+i)^{3}}{2+3 i}$

$=\frac{\left(4+i^{2}+4 i\right)(2+i)}{2+3 i} \quad\left(\because i^{2}=-1\right)$

$=\frac{8+2 i^{2}+8 i+4 i+i^{3}+4 i^{2}}{2+3 i}$

$=\frac{2+11 i}{2+3 i} \times \frac{2-3 i}{2-3 i}$

$=\frac{4-6 i+22 i-33 i^{2}}{4-9 i^{2}}$

$=\frac{37+16 i}{4+9}$

$=\frac{37}{13}+\frac{16}{13} i$

(vi) $\frac{(1+i)(1+\sqrt{3 i})}{1-i}$

$=\frac{(1+i)(1+\sqrt{3 i})}{1-i} \times \frac{1+i}{1+i}$

$=\frac{(1+\sqrt{3} i)\left(1+i^{2}+2 i\right)}{1-i^{2}} \quad\left(\because i^{2}=-1\right)$

$=\frac{(1+\sqrt{3} i) 2 i}{2}$

$=i(1+\sqrt{3} i)$

$=i+\sqrt{3} i^{2}$

$=-\sqrt{3}+i$

(vii) $\frac{2+3 i}{4+5 i}$

$=\frac{2+3 i}{4+5 i} \times \frac{4-5 i}{4-5 i}$

$=\frac{8-10 i+12 i-15 i^{2}}{16-25 i^{2}} \quad\left(\because i^{2}=-1\right)$

$=\frac{23+2 i}{16+25}$

$=\frac{23}{41}+\frac{2}{41} i$

(viii) $\frac{(1-i)^{3}}{1-i^{3}}$

$=\frac{\left(1+i^{2}-2 i\right)(1-i)}{1-i^{3}} \quad\left(\because i^{2}=-1\right)$

$=\frac{-2 i(1-i)}{1-i^{3}} \times \frac{1+i^{3}}{1+i^{3}}$

$=\frac{-2 i\left(1+i^{3}-i-i^{4}\right)}{1-i^{6}}$

$=\frac{-2 i(1-i-i-1)}{1-i^{2}}$

$=\frac{-2 i(-2 i)}{2}$

$=-2+0 i$

$(\mathrm{ix})(1+2 i)^{-3}$

$=\frac{1}{(1+2 i)^{3}}$

$=\frac{1}{1+8 i^{3}+6 i+12 i^{2}}$

$=\frac{1}{1-8 i+6 i-12} \quad\left(\because i^{2}=-1 \& i^{3}=-i\right)$

$=\frac{1}{-2 i-11}$

$=\frac{1}{-2 i-11} \times \frac{-2 i+11}{-2 i+11}$

$=\frac{-2 i+11}{4 i^{2}-121}$

$=\frac{-2 i+11}{-4-121}$

$=\frac{-2 i+11}{-125}$

$=-\frac{11}{125}+\frac{2 i}{125}$

$(\mathrm{x}) \frac{3-4 i}{(4-2 i)(1+i)}$

$=\frac{3-4 i}{4+2 i-2 i^{2}} \quad\left(\because i^{2}=-1\right)$

$=\frac{3-4 i}{6+2 i}$

$=\frac{3-4 i}{6+2 i} \times \frac{6-2 i}{6-2 i}$

$=\frac{18-6 i-24 i+8 i^{2}}{36-4 i^{2}}$

$=\frac{18-30 i-8}{36+4}$

$=\frac{10-30 i}{40}$

$=\frac{1}{4}-\frac{3}{4} i$

$(x i)\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{1+i-2+8 i}{(1-4 i)(1+i)}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{-1+9 i}{1-3 i+4 i^{2}}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{-1+9 i}{5-3 i}\right)\left(\frac{3-4 i}{5+i}\right) \quad\left(\because i^{2}=-1\right)$

$=\frac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}$

$=\frac{33+31 i}{28-10 i}$

$=\frac{33+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i}$

$=\frac{924+330 i+868 i+310 i^{2}}{784-100 i^{2}}$

$=\frac{614+1198 i}{884}$

$=\frac{307}{442}+\frac{599}{442} i$

(xii) $\frac{5+\sqrt{2} i}{1-\sqrt{2} i}$

$=\frac{5+\sqrt{2} i}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i}$

$=\frac{5+5 \sqrt{2} i+\sqrt{2} i+2 i^{2}}{1-2 i^{2}}$

$=\frac{5+6 \sqrt{2} i-2}{1+2} \quad\left(\because i^{2}=-1\right)$

$=\frac{3+6 \sqrt{2} i}{3}$

$=1+2 \sqrt{2} i$

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