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

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(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

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$0$

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(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

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