If z_{1} = 2 - i and z_{2} = 1 + i,find left| | vedclass

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

Question

(B) Given $z_{1} = 2 - i$ and $z_{2} = 1 + i$. First,calculate the numerator: $z_{1} + z_{2} + 1 = (2 - i) + (1 + i) + 1 = 4$. Next,calculate the deno

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) Given $z_{1} = 2 - i$ and $z_{2} = 1 + i$. First,calculate the numerator: $z_{1} + z_{2} + 1 = (2 - i) + (1 + i) + 1 = 4$. Next,calculate the denominator: $z_{1} - z_{2} + 1 = (2 - i) - (1 + i) + 1 = 2 - i - 1 - i + 1 = 2 - 2i$. Now,substitute these into the expression: $\left| \frac{4}{2 - 2i} \right| = \left| \frac{4}{2(1 - i)} \right| = \left| \frac{2}{1 - i} \right|$. To simplify,multiply the numerator and denominator by the conjugate $(1 + i)$: $\left| \frac{2(1 + i)}{(1 - i)(1 + i)} \right| = \left| \frac{2(1 + i)}{1^2 - i^2} \right|$. Since $i^2 = -1$,we have $\left| \frac{2(1 + i)}{1 + 1} \right| = \left| \frac{2(1 + i)}{2} \right| = |1 + i|$. The modulus is $|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

If $z_{1} = 2 - i$ and $z_{2} = 1 + i$,find $\left| \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} \right|$.

Similar Questions

If ${z_1}$ and ${z_2}$ are two complex numbers,then $|{z_1} + {z_2}|$ is

If $(1 + i) \cdot (1 + 2i) \dots \dots \dots (1 + ni) = x + iy$ (where $i = \sqrt{-1}$), then find the value of $(2) \cdot (5) \cdot (10) \dots \dots \dots (1 + n^2)$.

The real value of $\theta$ for which the expression $\frac{1 + i \cos \theta}{1 - 2i \cos \theta}$ is a real number is $(n \in I)$:

Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.

If $x + \frac{1}{x} = \sqrt{3},$ then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module