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 $(x-iy)(3+5i)$ is the conjugate of $-6-24i$ (where $x, y \in R$ and $i=\sqrt{-1}$),then the values of $x$ and $y$ are respectively:

If ${i^2} = -1$,then the sum $i + {i^2} + {i^3} + \dots$ up to $1000$ terms is equal to

Express the given complex number in the form $a+ib$: $\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$

If $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates,then $x=$

The value of $\sum_{n=1}^{13}(i^{n}+i^{n+1})$, where $i=\sqrt{-1}$, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module