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Question 6, Exercise 1.2
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t)\\ &=|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}\\ \Rightarrow \,\,\,\,|{{z}_{1}}{{z}_{2}}|&=|{{z}_{1}}||{{z}_{2
Question 8, Exercise 1.2
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===Question 8(iv)===== Show that $z=\overline{z}\Rightarrow z$ is real. ====Solution==== Suppose $z=a+bi$ .