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Mathematics Question on Complex Numbers and Quadratic Equations

Reduce (11−4i−21+i)(3−4i5+i)(\frac{1}{1-4i}-\frac{2}{1+i})(\frac{3-4i}{5+i}) to the standard form .

Answer

(\frac{1}{1-4i}-\frac{2}{1+i})(\frac{3-4i}{5+i})$$=[\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}][\frac{3-4i}{5+i}]

=[1+i−2+8i1+i−4i−4i2][3−4i5−3i]=[−1+9i5−3i][3−4i5+i]=[\frac{1+i-2+8i}{1+i-4i-4i^2}][\frac{3-4i}{5-3i}]=[\frac{-1+9i}{5-3i}][\frac{3-4i}{5+i}]

[−3+4i+27i−36225+5i−15i−3i2]=33+31i−4i28−10i=33+31i14−5i[\frac{-3+4i+27i-36^2}{25+5i-15i-3i^2}]=\frac{33+31i-4i}{28-10i}=\frac{33+31i}{14-5i}

=(33+3li)2(14−5i)×(14+5i)(14−5i)=\frac{(33+3li)}{2(14-5i)}×\frac{(14+5i)}{(14-5i)} [on multiplaying numerator and denominator by (14+5i)][on\,multiplaying\,numerator\,and\,denominator\,by\,(14+5i)]

462+165i+434i+155i22[(14)2(5i)2]=307+599i2(196−25i2)\frac{462+165i+434i+155i^2}{2[(14)^2(5i)^2]}=\frac{307+599i}{2(196-25i^2)}

=307+599i2(221)=307+599i442=307442+599i442=\frac{307+599i}{2(221)}=\frac{307+599i}{442}=\frac{307}{442}+\frac{599i}{442}

This is the required standard form.