Question

The value of $\left( z+3 \right)\left( \overline{z}+3 \right)$ is equivalent to:
(a)${{\left| z+3 \right|}^{2}}$
(b) $\left| z-3 \right|$
(c) ${{z}^{2}}+3$
(d) None of these

Answer 1

Hint: Taking $z=x+iy$ then $\overline{z}=x-iy$ and substituting in the expression $\left( z+3 \right)\left( \overline{z}+3 \right)$ and then solve it.

Complete step-by-step answer:
Now, substituting the value of $z$ and $\overline{z}$ in$\left( z+3 \right)\left( \overline{z}+3 \right)$, we get:

$\begin{align}

  & \left( x+iy+3 \right)\left( x-iy+3 \right) \\

 & \Rightarrow \left( x+iy \right)\left( x-iy \right)+\left( x+iy \right)3+3\left( x-iy \right)+9 \\

 & \\

\end{align}$

Now (x + iy)(x - iy) is in the form of (a + b)(a - b) identity so after using the identity we can come to an expression like x2 – (i)2y2 + 3x + 3iy + 3x -3iy + 9 and as we know i2 = -1 so we can write the expression as follows:

${{x}^{2}}+{{y}^{2}}+3x+3iy+3x-3iy+9$

Now, +3iy and -3iy will be cancelled out so the remaining expression will look as below:

${{x}^{2}}+{{y}^{2}}+6x+9$

We can rewrite the above expression as:

${{x}^{2}}+2.3.x+{{3}^{2}}+{{y}^{2}}$

As we can see a perfect square of (x + 3)2, so the above expression can be written as:

${{\left( x+3 \right)}^{2}}+{{y}^{2}}$

Now, we know that if $z=x+iy$then${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$. And${{\left| z+3 \right|}^{2}}={{\left( x+3 \right)}^{2}}+{{y}^{2}}$.As in modulus real term is written with x and imaginary term is clubbed with y. So the above expression can be written as:

${{\left| z+3 \right|}^{2}}$

Hence, the correct answer is option (a).

Note: There is an alternative way to solve the above problem. The expression $\left( z+3 \right)\left( \overline{z}+3 \right)$ is in the form of$z\overline{z}$. As $\left( \overline{z}+3 \right)$ is another way of writing $\left( \overline{z+3} \right)$ so we can also write as$\left( z+3 \right)\left( \overline{z+3} \right)$. And we know from the complex number chapter that $z\overline{z}={{\left| z \right|}^{2}}$ so $\left( z+3 \right)\left( \overline{z}+3 \right)={{\left| z+3 \right|}^{2}}$.Hence, we can directly use this property of complex number $z\overline{z}={{\left| z \right|}^{2}}$ to solve the above problem which would be a time saver in giving multiple choice question exams.