Question

Write $(1+2i)\cdot (1+3i)\cdot (2+i{)}^{-1}$ in the form $(a+ib)$ and Find the square roots of the given complex number $(7+24i)$.

Answer 1

Given,

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

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

$=\frac{-5+15i}{3}$

$=-\frac{5}{3}+3i$................$Ans.\left(i\right)$

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Let,

$7+24i=(a+ib)(a+ib)$

$7+24i=a^2-b^2+2abi$

$a^2-b^2=7$

$24=2ab\Rightarrow ab=12\Rightarrow a=\frac{12}{b}$

${\left(\frac{12}{b}\right)}^2-b^2=7$

$144-b^4=7b^2$

$b^4+7b^2-144=0$

$(b^2+16)(b^2-9)=0$

$\therefore b=3$

$a=\frac{12}{b}=\frac{12}{3}=4$

$\therefore z=4+3i$

$\therefore \sqrt{7+24i}=4+3i$................$Ans.\left(ii\right)$