The complex number frac{1 + 2i}{1 - i} lies i | vedclass

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(B) To find the quadrant,we first simplify the complex number $z = \frac{1 + 2i}{1 - i}$.Multiply the numerator and denominator by the conjugate of th

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(B) To find the quadrant,we first simplify the complex number $z = \frac{1 + 2i}{1 - i}$.Multiply the numerator and denominator by the conjugate of the denominator,which is $1 + i$:$z = \frac{1 + 2i}{1 - i} 	imes \frac{1 + i}{1 + i}$$z = \frac{1 + i + 2i + 2i^2}{1^2 - i^2}$Since $i^2 = -1$,we have:$z = \frac{1 + 3i - 2}{1 - (-1)} = \frac{-1 + 3i}{2} = -\frac{1}{2} + i\frac{3}{2}$The real part is $-\frac{1}{2}$ (negative) and the imaginary part is $\frac{3}{2}$ (positive).$A$ complex number with a negative real part and a positive imaginary part lies in the $II$ quadrant.

The complex number $\frac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane?

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