Solve for z
z=1+i
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z≔1+i
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z=i\left(1+\frac{i}{1i^{2}}\right)
Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.
z=i\left(1+\frac{i}{-1}\right)
By definition, i^{2} is -1. Calculate the denominator.
z=i\left(1-i\right)
Divide i by -1 to get -i.
z=i-i^{2}
Multiply i times 1-i.
z=i-\left(-1\right)
By definition, i^{2} is -1.
z=1+i
Do the multiplications in i-\left(-1\right). Reorder the terms.
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