In the Argand plane,the quadrant in which fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the quadrant,we first simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator,which i

Vedclass · Accepted Answer

Second

Vedclass · Answer

(B) To find the quadrant,we first simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator,which is $1+i$: $\frac{1+2i}{1-i} 	imes \frac{1+i}{1+i} = \frac{1+i+2i+2i^2}{1^2-i^2}$ Since $i^2 = -1$,we have: $\frac{1+3i-2}{1-(-1)} = \frac{-1+3i}{2} = -\frac{1}{2} + \frac{3}{2}i$ 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 Second quadrant.

In the Argand plane,the quadrant in which $\frac{1+2i}{1-i}$ lies is:

Similar Questions

If $(3x+2)-(5y-3)i$ and $(6x+3)+(2y-4)i$ are conjugates of each other,then the value of $\frac{x-y}{x+y}$ is (where $i=\sqrt{-1}, x, y \in R$ ).

The imaginary part of $\frac{(1-i)^3}{(2-i)(3-2i)}$ is

The least positive integer $n$ which will reduce $\left( \frac{i - 1}{i + 1} \right)^n$ to a real number is

The smallest positive integer $n$ for which $\frac{(1 + i)^n}{(1 - i)^{n-2}}$ is a real number, is ...

If $z=(1-i)^3(x+i)$ is a purely imaginary number for $x=x_1$ and if $z$ is a purely real number for $x=x_2$,then $x_1 x_2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module