Solve for z
z=8-4i
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z≔8-4i
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z=1\times 2+1\times \left(2i\right)-3i\times 2-3\times 2i^{2}
Multiply complex numbers 1-3i and 2+2i like you multiply binomials.
z=1\times 2+1\times \left(2i\right)-3i\times 2-3\times 2\left(-1\right)
By definition, i^{2} is -1.
z=2+2i-6i+6
Do the multiplications in 1\times 2+1\times \left(2i\right)-3i\times 2-3\times 2\left(-1\right).
z=2+6+\left(2-6\right)i
Combine the real and imaginary parts in 2+2i-6i+6.
z=8-4i
Do the additions in 2+6+\left(2-6\right)i.
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