Question

The curve represented by $\operatorname{Re} \left( {{z^2}} \right) = 4$ is
A. a parabola
B. an ellipse
C. a circle
D. a rectangular hyperbola

Answer 1

Hint: First, we will take $z = x + iy$ and then find the square of $z$. Then we will calculate the real part of the obtained equation to find the curve represented by the given equation.

Complete step by step answer :
Given that the equation $\operatorname{Re} \left( {{z^2}} \right) = 4$.

Let us assume that $z = x + iy$.

Squaring this equation on both sides, we get

$$\Rightarrow {z^2} = {\left( {x + iy} \right)^2} \\\ \Rightarrow {z^2} = {x^2} + {i^2}{y^2} + 2ixy \\\ \Rightarrow {z^2} = {x^2} - {y^2} + 2xyi \\\$$

Now we will find the real part of the above equation.

$$\Rightarrow \operatorname{Re} \left( {{z^2}} \right) = \operatorname{Re} \left( {{x^2} - {y^2} + 2xyi} \right) \\\ \Rightarrow \operatorname{Re} \left( {{z^2}} \right) = {x^2} - {y^2} \\\$$

Using the value $\operatorname{Re} \left( {{z^2}} \right) = 4$ in the above equation, we get

$$\Rightarrow 4 = {x^2} - {y^2} \\\ \Rightarrow {x^2} - {y^2} = 4 \\\ \Rightarrow \dfrac{{{x^2} - {y^2}}}{4} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{4} = 1 \\\$$

We know that the equation of hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$.

Thus, the above equation is an equation of a rectangular hyperbola.

Hence, the option D is correct.

Note: In this question, we will compare the solution with the general form of the equations for ellipse, circle, hyperbola and parabola to identify the curve. Also, we are supposed to write the values properly to avoid any miscalculation.