Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.3

Question 5

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Solve the following equations .

(ii) 3z5243z=03z^5 - 243z = 0

Solution

Factor out common factor 3z:3z(z481)=0Apply zero product property:3z=0    z=0z481=0    z4=81Difference of squares:(z29)(z2+9)=0z29=0    z2=9    z=±3z2+9=0    z2=9    z=±3iTherefore: z=0,3,3,3i,3i\begin{aligned} & \boxed{\text{Factor out common factor 3z:}} \\ \\ & 3z(z^4 - 81) = 0 \\ \\ & \boxed{\text{Apply zero product property:}} \\ \\ & 3z = 0 \implies z = 0 \\ \\ & z^4 - 81 = 0 \implies z^4 = 81 \\ \\ & \boxed{\text{Difference of squares:}} \\ \\ & (z^2 - 9)(z^2 + 9) = 0 \\ \\ & z^2 - 9 = 0 \implies z^2 = 9 \implies z = \pm 3 \\ \\ & z^2 + 9 = 0 \implies z^2 = -9 \implies z = \pm 3i \\ \\ & \boxed{\text{Therefore: } z = 0, 3, -3, 3i, -3i} \end{aligned}