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

asked by guest
on Mar 30, 2025 at 6:11 am



You asked:

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

MathBot Answer:

The solution is: Z=−3+7iZ = -3 + 7 i

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

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

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