Answer to Question #347724 in Analytic Geometry for Merryza

Question

Given z1=2<45degrees, z2=3<120degrees, z3=4<180degrees. Determine the following:

a) (z1)^2+z2/z2+z3

b) z1/z2*z3

Answer 1

(a)

$$(z_1)^2=(2)^2\angle(2\cdot45\degree)=4i$$ $$z_2+z_3=3(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i)-4=-\dfrac{11}{2}-\dfrac{3\sqrt{3}}{2}i)$$ $$\dfrac{z_2}{z_2+z_3}=\dfrac{-3+3\sqrt{3}i}{-11-3\sqrt{3}i}$$ $$=\dfrac{(-3+3\sqrt{3}i)(-11-3\sqrt{3}i)}{121+27}$$ $$=\dfrac{33+9\sqrt{3}i-33\sqrt{3}i+27}{148}$$ $$=\dfrac{15}{37}-\dfrac{6\sqrt{3}}{37}i$$ $$(z_1)^2+\dfrac{z_2}{z_2+z_3}=4i+\dfrac{15}{37}-\dfrac{6\sqrt{3}}{37}i$$ $$=\dfrac{15}{37}+\dfrac{148-6\sqrt{3}}{37}i$$

(b)

$$z_2z_3=3(4)\angle(120\degree+180\degree)=6-6\sqrt{3}i$$ $$\dfrac{z_1}{z_2z_3}=\dfrac{\sqrt{2}+\sqrt{2}i}{6-6\sqrt{3}i}$$ $$=\dfrac{(\sqrt{2}+\sqrt{2}i)(6+6\sqrt{3}i)}{36+108}$$ $$=\dfrac{6\sqrt{2}+6\sqrt{6}i+6\sqrt{2}i-6\sqrt{6}}{144}$$ $$=-\dfrac{\sqrt{6}-\sqrt{2}}{24}+\dfrac{\sqrt{6}+\sqrt{2}}{24}i$$