z=2-2i-2-6i-3-15i
asked by guest on Mar 30, 2025 at 6:08 am
You asked:
MathBot Answer:
The solution is: z=−3−23iz = -3 - 23 iz=−3−23i
z=2−2i−2−6i−3−15iz=−3−23i\begin{aligned}z &= 2 - 2 i - 2 - 6 i - 3 - 15 i\\z &= -3 - 23 i\end{aligned}zz=2−2i−2−6i−3−15i=−3−23i
iii is the imaginary unit, defined as i2=−1i^2 = -1i2=−1.
