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Miscellaneous Exercise 1 (Subjective) · Q192

Q.Show that z=5(1−i)(2−i)(3−i)z = \dfrac{5}{(1-i)(2-i)(3-i)} is purely imaginary number.

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(1−i)(2−i)=2−i−2i+i2=2−3i−1=1−3i(1-i)(2-i)=2-i-2i+i^2=2-3i-1=1-3i. Then (1−3i)(3−i)=3−i−9i+3i2=3−10i−3=−10i(1-3i)(3-i)=3-i-9i+3i^2=3-10i-3=-10i. So z=dfrac5−10i=dfrac5−10itimesdfracii=dfrac5i−10i2=dfrac5i10=dfraci2z=\\dfrac{5}{-10i}=\\dfrac{5}{-10i}\\times\\dfrac{i}{i}=\\dfrac{5i}{-10i^2}=\\dfrac{5i}{10}=\\dfrac{i}{2}. This has real part $ …

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