Solvetheequationsbelowandhence,showthesolutionsoftheseequationsontheArganddiagram:(i)z4−i=0(ii)z3=−√3−i(iii)z5=−18i(iv)z=16i1+i18(v)z=1+i√3+i16
(i)z4−i=0z4-i=0z4−i=0
4(x+iy)−i=0⇒4x+(4y−1)i=0⇒x=0 and y=144(x+iy)-i=0\Rightarrow 4x+(4y-1)i=0\Rightarrow x=0 \text{ and }y=\dfrac{1}{4}4(x+iy)−i=0⇒4x+(4y−1)i=0⇒x=0 and y=41
(ii) z3=−3−iz3=-\sqrt{3}-iz3=−3−i
3(x+iy)+3+i=0⇒(3x+3)+i(y+1)=0⇒x=−13,y=−13(x+iy)+\sqrt{3}+i=0\Rightarrow (3x+\sqrt{3})+i(y+1)=0\Rightarrow x=\dfrac{-1}{\sqrt{3}}, y=-13(x+iy)+3+i=0⇒(3x+3)+i(y+1)=0⇒x=3−1,y=−1
(iii)z5=−18iz5=-18iz5=−18i
5(x+iy)+18i=0⇒5x+i(y+18)=0⇒x=0,y=−185(x+iy)+18i=0\Rightarrow 5x+i(y+18)=0\Rightarrow x=0,y=-185(x+iy)+18i=0⇒5x+i(y+18)=0⇒x=0,y=−18
(iv) z=16i+18iz=16i+18iz=16i+18i
x+iy=34i⇒x=0,y=34x+iy=34i\Rightarrow x=0,y=34x+iy=34i⇒x=0,y=34
(v) z=1+i3+16iz=1+i\sqrt{3}+16iz=1+i3+16i
x+iy=1+i(3+16)⇒x=1,y=3+16x+iy=1+i(\sqrt{3}+16)\Rightarrow x=1,y=\sqrt{3}+16x+iy=1+i(3+16)⇒x=1,y=3+16
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