Find the conjugates of the following complex numbers:
Question:

Find the conjugates of the following complex numbers:

(i) $4-5 i$

(ii) $\frac{1}{3+5 i}$

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

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

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

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

Solution:

(i) Let $z=4-5 i$

$\therefore \bar{z}=4+5 i \quad(z=a+i b$, so $\bar{z}=a-i b)$

(ii) Let $z=\frac{1}{3+5 i}$

$=\frac{1}{3+5 i} \times \frac{3-5 i}{3-5 i}$

$=\frac{3-5 i}{9-25 i^{2}}$

$=\frac{3-5 i}{9+25}$

$=\frac{3-5 i}{34}$

$\therefore \bar{z}=\frac{3+5 i}{34}$

(iii) Let $z=\frac{1}{1+i}$

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

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

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

$\Rightarrow \bar{z}=\frac{1+i}{2}$

(iv) Let $z=\frac{(3-i)^{2}}{2+i}$

$=\frac{(9-6 i-1)}{2+i}$

$=\frac{8-6 i}{2+i} \times \frac{2-i}{2-i}$

$=\frac{16-8 i-12 i+6 i^{2}}{4-i^{2}}$

$=\frac{10-20 i}{5}$

$=2-4 i$

$\therefore z=2+4 i$

(v) Let $z=\frac{(1+i)(2+i)}{3+i}$

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

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

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

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

$=\frac{6+8 i}{10}$

$=\frac{3+4 i}{5}$

$\Rightarrow \bar{z}=\frac{3-4 i}{5}$

(vi) Let $z=\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$

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

$=\frac{6+6+5 i}{2+2+3 i}$

$=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i}$

$=\frac{48-36 i+20 i-15 i^{2}}{16-9 i^{2}}$

$=\frac{63-16 i}{25}$

$\therefore \bar{z}=\frac{63+16 i}{25}$

 

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