z=2i-2+2*(-1-3i)-3*(2-2i-1-3i)

asked by guest
on Mar 30, 2025 at 6:09 am



You asked:

Investigate the equation: z=2i−2+2(−1−3i)−3(2−2i−1−3i)z = 2 i - 2 + 2 \left(-1 - 3 i\right) - 3 \left(2 - 2 i - 1 - 3 i\right).

MathBot Answer:

The solution is: z=−7+11iz = -7 + 11 i

z=2i−2+2(−1−3i)−3(2−2i−1−3i)z=−7+11i\begin{aligned}z &= 2 i - 2 + 2 \left(-1 - 3 i\right) - 3 \left(2 - 2 i - 1 - 3 i\right)\\z &= -7 + 11 i\end{aligned}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

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