Question

Find the conjuagates of the following complex numbers :

$\left(i\right)4-5i\left(ii\right)\frac{1}{3+5i}\left(iii\right)\frac{1}{1+i}\left(iv\right)\frac{(3-i{)}^2}{2+i}\left(v\right)\frac{(1+i)(2+i)}{3+i}\left(vi\right)\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

Answer 1

(i) 4 - 5i

If z = x + iy is a complex number, then the conjugate of z denoted by $\overline{z}$ is defined as $\overline{z}$ = x - iy

let z = 4 - 5i

$\Rightarrow \overline{z}$ = 4 + 5 i

$\left(ii\right)\frac{1}{3+5i}Letz=\frac{1}{3+5i}=\frac{1}{3+5i}\times \frac{(3-5i)}{3-5i}=\frac{3-5i}{3^2+5^2}\Rightarrow z=\frac{3-5i}{9+25}So\overline{z}=\frac{3+5i}{34}=\frac{3}{34}+\frac{5}{34}i$

$\left(iii\right)\frac{1}{1+i}Letz=\frac{1}{1+i}=\frac{1}{1+i}\times \frac{(1-i)}{(1-i)}=\frac{1-i}{1^2+1^1}=\frac{1-i}{2}\therefore \overline{z}=\frac{1+i}{2}=\frac{1}{2}+\frac{1}{2}i$

$\left(iv\right)\frac{(3-i{)}^2}{2+i}Letz=\frac{(3-i{)}^2}{2+i}=\frac{3^2+i^2-2\times 3\times i}{2+i}=\frac{9-1-6i}{2+i}=\frac{8-6i}{2+i}=\frac{8-6i}{2+i}\times \frac{2-i}{2-i}=\frac{8(2-i)-6i(2-i)}{2^2+1^2}=\frac{16-8i-12i-6}{4+1}=\frac{10-2i}{5}\Rightarrow z=2-4i\text{Hence,}\overline{z}=2+4i$

$\left(v\right)\frac{(1+i)(2+i)}{3+i}Letz=\frac{(1+i)(2+i)}{3+i}=\frac{2+i+i(2+i)}{3+i}=\frac{2+i+2i-1}{3+i}=\frac{1+3i}{3+i}=\frac{(1+3i)}{3+i}\times \frac{(3-i)}{(3-i)}=\frac{3-i+3i(3-i)}{3^2+1^2}=\frac{3-i+9i+3}{9+1}=\frac{6+8i}{10}=\frac{2(3+4i)}{10}\Rightarrow z=\frac{3+4i}{5}Hence\overline{z}=\frac{3-4i}{5}\frac{3}{5}-\frac{4}{5}i$

$\left(vi\right)\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}Letz=\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}=\frac{3(2+3i)-2i(2+3i)}{2-i+2i(2-i)}=\frac{6+9i-4i+6}{2-i+4i+2}=\frac{12+5i}{4+3i}=\frac{12+5i}{4+3i}\times \frac{4-3i}{4-3i}=\frac{12(4-3i)+5i(4-3i)}{4(4-3i)+3i(4-3i)}=\frac{48-36i+20i+15}{16-12i+12i+9}=\frac{63-16i}{16+9}\Rightarrow z=\frac{63-16i}{25}\therefore \overline{z}=\frac{63+16i}{25}=\frac{63}{25}+\frac{16}{25}i$