Question

Z=1+3i/1-2i in a+ib form
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Answer 1

Answer:

Z = $\frac {1+3i}{1-2i}$

Z = $\frac {1+3i}{1-2i}$ * $\frac {1+2i}{1+2i}$

Z = $\frac {1 + 2i + 3i + 6i^2}{1^2 - (2i)^2}$

Z = $\frac {1 - 6 + 5i}{1 - 4i^2}$

Z = $\frac {-5 + 5i}{1-4(-1)}$

Z = $\frac {5(-1 + i)}{5}$

Z = -1 + i.

here a = -1 & b = 1

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