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Question 6, Exercise 1.2 @fsc-part1-kpk:sol:unit01

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1}}||{{z}_{2}}|=R.H.S. \end{align} **Alternative Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{a

Question 8, Exercise 10.1 @fsc-part1-kpk:sol:unit10

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\right)}\\ &=R.H.S.\end{align} ===Alternative Method=== \begin{align}L.H.S.&=\tan\left( \dfrac{\pi }{4