Question

Write 6i^5+ 7i^4 + 3i^3 + 5i^2-4 in the form of a +ib and find its square root.​

Answer 1

Answer:-

$\sf{Hence, \ the \ square \ root \ of }$

$\sf{6i^{5}+7i^{4}+3i^{3}+5i^{2}-4 \ is }$

$\sf{\sqrt3+\frac{\sqrt3}{2}i \ or \ -3-\frac{\sqrt3}{2}i}$

Given:

$\sf{\implies{6i^{5}+7i^{4}+3i^{3}+5i^{2}-4}}$

Solution:

$\sf{\implies{6i^{5}+7i^{4}+3i^{3}+5i^{2}-4}}$

$\sf{\implies{6(i^{4}.i)+7i^{4}+3i^{3}+5i^{2}-4}}$

$\sf\blue{\because{i^{2}=-1, \ i^{3}=-i \ and \ i^{4}=1}}$

$\sf{\implies{6i+7-3i-5-4}}$

$\sf{\implies{-2+3i}}$

$\sf{Let \ \sqrt{-2+3i}=a+bi, \ where \ a \ and \ b \ \in \ R}$

$\sf{Squaring \ both \ sides, \ we \ get}$

$\sf{-2+3i=(a+bi)^{2}}$

$\sf{\therefore{-2+3i=a^{2}+b^{2}i^{2}+2abi}}$

$\sf{\therefore{-2+3i=(a^{2}-b^{2})+2abi...[\because{i^{2}=-1}]}}$

$\sf{Equating \ real \ and \ imaginary \ parts,}$

$\sf{we \ get}$

$\sf{a^{2}-b^{2}=-2 \ and \ 2ab=3}$

$\sf{\therefore{a^{2}-b^{2}=-2 \ and \ b=\frac{3}{2a}}}$

$\sf{\therefore{a^{2}-(\frac{3}{2a})^{2}=-2}}$

$\sf{\therefore{a^{2}-\frac{9}{4a^{2}}=-2}}$

$\sf{\therefore{\frac{4a^{4}-9}{4a^{2}}=-2}}$

$\sf{\therefore{4a^{4}-9=8a^{2}}}$

$\sf{\therefore{4a^{2}-8a^{2}-9=0}}$

$\sf{\therefore{4a^{2}-12a^{2}+3a^{2}-9=0}}$

$\sf{\therefore{4a^{2}(a^{2}-3)+3(a^{2}-3)=0}}$

$\sf{\therefore{(a^{2}-3)(4a^{2}+3)=0}}$

$\sf{\therefore{a^{2}=3 \ or \ a^{2}=\frac{-3}{4}}}$

$\sf{But, \ a \ \in \ R}$

$\sf{\therefore{a^{2}\ \neq \ \frac{-3}{4}}}$

$\sf{\therefore{a^{2}=3}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{a=\pm\sqrt3}$

$\sf{When \ a=\sqrt3, \ b=\frac{3}{(2)(\sqrt3)}}$

$\sf{\therefore{b=\frac{\sqrt3}{2}}}$

$\sf{When \ a=-\sqrt3, \ b=\frac{3}{(2)(-\sqrt3)}}$

$\sf{\therefore{b=\frac{-\sqrt3}{2}}}$

$\sf{\therefore{a+bi=\pm(\sqrt3+\frac{\sqrt3}{2}i)}}$

$\sf{\therefore{\sqrt{-2+3i}=\pm(\sqrt3+\frac{\sqrt3}{2}i)}}$

$\sf\purple{\tt{\therefore{Hence, \ the \ square \ root \ of }}}$

$\sf\purple{\tt{6i^{5}+7i^{4}+3i^{3}+5i^{2}-4 \ is }}$

$\sf\purple{\tt{\sqrt3+\frac{\sqrt3}{2}i \ or \ -3-\frac{\sqrt3}{2}i}}$