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Miscellaneous Exercise 1 (Subjective) · Q205

Q.Simplify 11−2i+31+i+3+4i2−4i\dfrac{1}{1-2i}+\dfrac{3}{1+i}+\dfrac{3+4i}{2-4i}

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dfrac11−2i=dfrac1+2i5\\dfrac{1}{1-2i}=\\dfrac{1+2i}{5}. dfrac31+i=dfrac3(1−i)2=dfrac3−3i2\\dfrac{3}{1+i}=\\dfrac{3(1-i)}{2}=\\dfrac{3-3i}{2}. dfrac3+4i2−4i=dfrac(3+4i)(2+4i)(2−4i)(2+4i)=dfrac6+12i+8i+16i24+16=dfrac6+20i−1620=dfrac−10+20i20=−dfrac12+i\\dfrac{3+4i}{2-4i}=\\dfrac{(3+4i)(2+4i)}{(2-4i)(2+4i)}=\\dfrac{6+12i+8i+16i^2}{4+16}=\\dfrac{6+20i-16}{20}=\\dfrac{-10+20i}{20}=-\\dfrac12+i. Sum all three over a common denominator 1010: $\dfrac{2(1+2i)}{10}+\dfrac{15(1-i)}{10}+\dfrac{10(-\tfrac12+i)}{10}=\dfrac{(2+4i)+(15-15i)+(- …

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