Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.3

Question 5

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Solve the following equations .

(i) 2z432=02z^4 - 32 = 0

Solution

Factor out common factor 2:2(z416)=0    z416=0Difference of squares: a2b2=(ab)(a+b)(z24)(z2+4)=0Apply zero product property:z24=0orz2+4=0z2=4    z=±2z2=4    z=±2iTherefore: z=2,2,2i,2i\begin{aligned} & \boxed{\text{Factor out common factor 2:}} \\ \\ & 2(z^4 - 16) = 0 \implies z^4 - 16 = 0 \\ \\ & \boxed{\text{Difference of squares: } a^2 - b^2 = (a - b)(a + b)} \\ \\ & (z^2 - 4)(z^2 + 4) = 0 \\ \\ & \boxed{\text{Apply zero product property:}} \\ \\ & z^2 - 4 = 0 \quad \text{or} \quad z^2 + 4 = 0 \\ \\ & z^2 = 4 \implies z = \pm 2 \\ \\ & z^2 = -4 \implies z = \pm 2i \\ \\ & \boxed{\text{Therefore: } z = 2, -2, 2i, -2i} \end{aligned}