The real part of the complex number z = frac{ | vedclass

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

Question

(B) First,simplify each term individually:$1$. $\frac{5+2i}{2-5i} = \frac{(5+2i)(2+5i)}{(2-5i)(2+5i)} = \frac{10 + 25i + 4i + 10i^2}{4 + 25} = \frac{1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) First,simplify each term individually:$1$. $\frac{5+2i}{2-5i} = \frac{(5+2i)(2+5i)}{(2-5i)(2+5i)} = \frac{10 + 25i + 4i + 10i^2}{4 + 25} = \frac{10 + 29i - 10}{29} = \frac{29i}{29} = i$$2$. $\frac{3-4i}{4+3i} = \frac{(3-4i)(4-3i)}{(4+3i)(4-3i)} = \frac{12 - 9i - 16i + 12i^2}{16 + 9} = \frac{12 - 25i - 12}{25} = \frac{-25i}{25} = -i$$3$. $\frac{1}{i} = \frac{1 	imes i}{i 	imes i} = \frac{i}{-1} = -i$Substituting these back into the expression for $z$:$z = (i) - (-i) - (-i) = i + i + i = 3i$The complex number is $z = 0 + 3i$.Thus,the real part of $z$ is $0$.

The real part of the complex number $z = \frac{5+2i}{2-5i} - \frac{3-4i}{4+3i} - \frac{1}{i}$ is

Similar Questions

If $\frac{2+3 i}{i-2}-\frac{4 i-3}{3+4 i}=x+i y$,then $3 x+y=$

If $(x + iy)(p + iq) = (x^2 + y^2)i$,then

Express the given complex number in the form $a+ib$: $i^{-39}$

If $4x + i(3x - y) = 3 + i(-6)$,where $x$ and $y$ are real numbers,then find the values of $x$ and $y$.

If $x + \frac{1}{x} = 2\cos \theta ,$ then $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module