Question

If we have $\operatorname{Re}\left( {{z}^{2}} \right)=0$ then the locus of z is.
$\begin{align}
  & \text{a) Circle} \\
 & \text{b) Parabola} \\
 & \text{c) Pair of non perpendicular lines} \\
 & \text{d) Pair of perpendicular lines} \\
\end{align}$

Answer 1

Hint: First we will assume z = x + iy. Then squaring on both sides we will simplify the equation using ${{\left( a+b \right)}^{2}}=\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$ Now we will equate the real part to 0 and hence find the required condition for locus.

Complete step-by-step solution:
Now let us say z = x + iy. where x and y are real numbers and ${{i}^{2}}=-1$
Hence squaring on both sides we get
${{z}^{2}}={{\left( x+iy \right)}^{2}}$
Now we know that the formula for ${{\left( a+b \right)}^{2}}$ is given by ${{\left( a+b \right)}^{2}}=\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$
Hence using this formula to evaluate RHS of our equation we get.
${{z}^{2}}=\left( {{x}^{2}}+2xyi+{{\left( iy \right)}^{2}} \right)$
Now we know that ${{\left( ab \right)}^{2}}={{a}^{2}}{{b}^{2}}$ hence we get
${{z}^{2}}=\left( {{x}^{2}}+2xyi+{{i}^{2}}{{y}^{2}} \right)$
Now substituting the value ${{i}^{2}}=-1$ in the above equation we get
$\begin{align}
  & {{z}^{2}}={{x}^{2}}+2xyi-{{y}^{2}} \\
 & \Rightarrow {{z}^{2}}=\left( {{x}^{2}}-{{y}^{2}} \right)+\left( 2xy \right)i \\
\end{align}$
Now in a complex number of form $ x + iy $ we say that a is the real part and b is the imaginary part.
Here we have 2ab multiplied with i hence 2ab is imaginary part of ${{z}^{2}}$and ${{a}^{2}}-{{b}^{2}}$ is real part of ${{z}^{2}}$
Now we have that the real part of ${{z}^{2}}$ is 0.
Hence we get
${{x}^{2}}-{{y}^{2}}=0$
Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Hence we get
$(y – x)(y + x) = 0$
Now if we have two real numbers a and b such that $ab = 0$ then $a = 0$ or $b = 0$.
Hence we get wither $y – x = 0$ or $y + x = 0$
Now we know that this is nothing but product of two straight lines y – x = 0 and y + x = 0.
Now we know that the slope of line $y = mx$ is m. hence the slope of line $y – mx$ is m
Hence we say that the slope of line $y – x = 0$ is 1 and slope of line $y + x = 0$ is -1.
Hence we can say that the lines are perpendicular
Hence z represents a pair of straight lines.

Note: Now note that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ but $\left( a+ib \right)={{a}^{2}}-{{b}^{2}}+2aib$ do not mistake by writing \[{{\left( a+ib \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2abi\] the minus sign arises because \[{{i}^{2}}=-1\]. The important thing is that after doing operation on complex number separate the real part and
imaginary part.