Question:

If z =$\frac{2-i}{i}$ = , then Re(z$^2$) + lm(z$^2$) is equal to

Updated On: Apr 8, 2024
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The Correct Option is A

Solution and Explanation

$ z=\frac{2-i}{i} \Rightarrow z^{2}= \frac{(2-i)^{2}}{i^{2}} $
$=\frac{4-4 i-1}{-1}=\frac{3-4 i}{-1}=4 i-3$
Now, $ \operatorname{Re}\left(z^{2}\right) =-3$ and $ \operatorname{lm}\left(z^{2}\right)=4$
$\therefore \operatorname{Re}\left(z^{2}\right)+\operatorname{lm}\left(z^{2}\right) =-3+4=1$
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.