Question

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

$\left(i\right)(1+i)(1+2i)\left(ii\right)\frac{3+2i}{-2+i}\left(iii\right)\frac{1}{(2+i{)}^2}\left(iv\right)\frac{1-i}{1+i}\left(v\right)\frac{(2+i{)}^3}{2+3i}\left(vi\right)\frac{(1+i)(1+\sqrt{3i})}{1-i}\left(vii\right)\frac{2+3i}{4+5i}\left(viii\right)\frac{(1-i{)}^3}{1-i^3}\left(ix\right)(1+2i{)}^{-3}(x\left)\frac{3-4i}{(4-2i)(1+i)}\right(xi\left)(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)\right(xii)\frac{5+\sqrt{2i}}{1-\sqrt{2i}}$

Answer 1

$\left(i\right)(1+i)(1+2i)=1\times (1+2i)+i(1+2i)=1+2i+i+2i^2=1+3i-2=-1+3i\therefore (1+i)(1+2i)=-1+3i$

$\left(ii\right)\frac{3+2i}{-2+i}=\frac{3+2i}{(-2+i)}\times \frac{-2-i}{-2-i}\text{[Rationalising the denominator]}=\frac{3(-2-i)+2i(-2-i)}{(-2{)}^2-(i{)}^2}[\because (a+ib\left)\right(a-ib)=a^2+b^2]=\frac{-6-3i-4i+2}{4+1}[\because -i^2=1]=\frac{-4-7i}{5}=\frac{-4}{5}-\frac{7}{5}i\therefore \frac{3+2i}{-2+i}=\frac{-4}{5}-\frac{7}{5}i$

$\left(iii\right)\frac{1}{(2+i{)}^2}=\frac{1}{(2+i{)}^2}=\frac{1}{2^2+(i{)}^2+2\times 2\times i}=\frac{1}{4-1+4i}=\frac{1}{3+4i}=\frac{1}{(3+4i)}\times \frac{(3-4i)}{(3-4i)}\text{[on rationalising the denominator]}=\frac{3-4i}{3^2+4^2}[\because (a+ib\left)\right(a-ib)=a^2+b^2]=\frac{3-4i}{25}=\frac{3}{25}-\frac{4}{25}i\therefore \frac{1}{(2+i{)}^2}=\frac{3}{25}-\frac{4}{25}i$

$\left(iv\right)\frac{1-i}{1+i}=\frac{(1-i)}{(1+i)}\times \frac{(1-i}{(1-i}\text{(Rationalising the denominator)}=\frac{(1-i{)}^2}{1^2+1^2}[\because (a+ib\left)\right(a-ib)=a^2+b^2]=\frac{1^2+i^2=2\times i\times 1}{2}=\frac{-2i}{2}=-i=0-i\therefore \frac{1-i}{1+i}=0-i$

$\left(v\right)\frac{(2+i{)}^3}{2+3i}=\frac{2^3+i^3+3\times 2\times i(2+i))}{2+3i}[\because (a+b{)}^3=a^3+b^3+3ab(a+b)]=\frac{(8-i+6i(2+i\left)\right)}{2+3i}\times \frac{(2-3i)}{2-3i}\text{(On rationalising the denominator)}=\frac{(8-i+12i+6i^2)(2-3i)}{2^2+3^2}=\frac{(8-6+11i)(2-3i)}{4+9}(\because i^2=-1)=\frac{(2+11i)(2-3i)}{13}=\frac{4-6i+22i+33}{13}=\frac{37+16i}{13}=\frac{37}{13}+\frac{16}{13}i\therefore \frac{(2+i{)}^3}{2+3i}=\frac{37}{13}+\frac{16}{13}i$

$\left(vi\right)\frac{(1+i)(1+\sqrt{3i})}{1-i}=\frac{1(1+\sqrt{3}+i(1+\sqrt{3i}\left)\right)}{1-i}=\frac{(1+\sqrt{3i}+i-\sqrt{3})}{1-i}(\because i^2=-1)=\frac{(1-\sqrt{3})+i(1+\sqrt{3})}{1-i}\times \frac{(1+i)}{1+i}\text{(Rationalising the denominator)}=\frac{(1-\sqrt{3})(1+i)+i(1+\sqrt{3})(1+i)}{1^2+1^2}=\frac{1+i-\sqrt{3}(1+i)+i(1+i+\sqrt{3}(1+i\left)\right)}{2}=\frac{1+i-\sqrt{3}-\sqrt{3}i+(1+i+\sqrt{3}+\sqrt{3}i)}{2}=\frac{1-\sqrt{3}+i-\sqrt{3i}-1+\sqrt{3i}-\sqrt{3}}{2}=\frac{-2\sqrt{3}+2i}{2}=-\sqrt{3}+i\therefore \frac{(1+i)(1+\sqrt{3i})}{1-i}=-\sqrt{3}+i$

$\left(vii\right)\frac{2+3i}{4+5i}=\frac{2+3i}{4+5i}\times \frac{(4-5i)}{(4-5i)}\text{(Rationalising the denominator)}=\frac{2(4-5i)+3i(4-5i)}{4^2+5^2}=\frac{8-10i+12i+15}{16+25}(\because i^2=-1)=\frac{23+2i}{41}=\frac{23}{41}+\frac{2}{41}i\therefore \frac{2+3i}{4+5i}=\frac{23}{41}+\frac{2}{41}i$

$\left(viii\right)\frac{(1-i{)}^3}{1-i^3}=\frac{1^3-i^3-3\times 1\times i(1-i)}{1-(-i)}[\because (a-b{)}^3=a^3-b^3-3ab(a-b)and{i}^3=-i]=\frac{1-(-i)-3i(1-i)}{1+i}=\frac{1+i-3i-3}{1+i}=\frac{-2-2i}{1+i}=\frac{-2(1+i)}{1+i}=-2=-2+0i\therefore \frac{(1+i{)}^3}{1-i^3}=-2+0i$

$\left(ix\right)(1+2i{)}^{-3}=\frac{1}{(1+2i{)}^3}(\because z^{-3}=\frac{1}{z^3})=\frac{1}{1^3+(2i{)}^3+3\times 1\times 2i(1+2i)}=\frac{1}{1^3+2^3\times i^3+6i(1+2i)}=\frac{1}{1-8i+6i-12}(\because i^3=-i\text{and}{i}^2=-1)=\frac{1}{-11-2i}=\frac{1}{-11-2i}\times \frac{(-11+2i)}{(-11+2i)}=\frac{-11+2i}{(-11{)}^2+2^2}=\frac{-11+2i}{121+4}=\frac{-11}{125}+\frac{2}{125}i\therefore (1+2i{)}^{-3}=\frac{-11}{125}+\frac{2}{125}i$

$\left(x\right)\frac{3-4i}{(4-2i)(1+i)}=\frac{3-4i}{4(1+i)-2i(1+i)}=\frac{3-4i}{4+4i-2i+2}=\frac{3-4i}{6+2i}=\frac{3-4i}{6+2i}\times \frac{6-2i}{6-2i}=\frac{3(6-2i)-4i(6-2i)}{6^2+2^2}=\frac{18-6i-24i-8}{36+4}=\frac{10-30i}{40}=\frac{10(1-3i)}{40}=\frac{1-3i}{4}=\frac{1}{4}-\frac{3}{4}i\therefore \frac{3-4i}{(4-2i)(1+i)}=\frac{1}{4}-\frac{3}{4}i$

$\left(xi\right)(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)=\frac{1+i-2(1-4i)}{(1-4i)(1+i)}\times \frac{3-4i}{5+i}=\frac{1+i-2+8i}{1(1+i)-4i(1+i)}\times \frac{3-4i}{5+i}=\frac{-1+9i}{(1+i-4i+4)}\times \frac{3-4i}{5+i}=\frac{-1(3-4i)+9i(3-4i)}{(5-3i)(5+i)}=\frac{-3+4i+27i+36}{5(5+i)-3i(5+i)}=\frac{33+31j}{25+5i-15i+3}=\frac{33+31i}{28-10j}=\frac{33+31i}{28-10i}\times \frac{(28+10i)}{28+10i}=\frac{33\times 28+33\times 10i+31i\times 28+31i\times 10i}{{28}^2+{10}^2}=\frac{924+330i+868i-310}{784+100}=\frac{614+1198i}{884}=\frac{614}{884}+\frac{1198}{884}i=\frac{307}{442}+\frac{599}{442}i\therefore (\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)=\frac{307}{442}+\frac{499}{442}i$

$\left(xii\right)\frac{5+\sqrt{2}i}{1-\sqrt{2}i}=\frac{5+\sqrt{2}i}{1-\sqrt{2}i}\times \frac{1+\sqrt{2}i}{1+\sqrt{2}i}=\frac{5(1+\sqrt{2}i)+\sqrt{2}i(1+\sqrt{2}i)}{1+2}=\frac{5+5\sqrt{2}i+\sqrt{2}i-2}{3}=\frac{3+6\sqrt{2}i}{3}=1+2\sqrt{2}i\text{Therefore,~}\frac{5+\sqrt{2}i}{1-\sqrt{2}i}=1+2\sqrt{2}i$