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Find the root of a complex number $5+12i$.

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Hint: We have to find the root of the complex number $5+12i$. Consider the root of $5+12i$ as $a+ib$, and solve it, which will lead you to find the value of $a$ and $b$. You will get the answer.

Complete step-by-step answer:
Consider the following example, which follows from basic algebra:
${{\left( 5{{e}^{3j}} \right)}^{2}}=25{{e}^{6j}}$

We can generalize this example as follows:
${{\left( r{{e}^{\theta j}} \right)}^{n}}={{r}^{n}}{{e}^{jn\theta }}$
The above expression, written in polar form, leads us to DeMoivre's Theorem.

DeMoivre's Theorem gives a formula for computing powers of complex numbers.

In mathematics, de Moivre's formula (also known as de Moivre's theorem and de Moivre's identity) states that for any real number $x$ and integer $n$ it holds that
${{(\cos (x)+i\sin (x))}^{n}}=\cos (nx)+i\sin (nx)$
where$i$is the imaginary unit (${{i}^{2}}=-1$)? The formula is named after Abraham de Moivre, although he never stated it in his works. The expression $\cos x+i\sin x$is sometimes abbreviated to $c is x$.

The formula is important because it connects complex numbers and trigonometry. By expanding the left-hand side and then comparing the real and imaginary parts under the assumption that $x$ is real, it is possible to derive useful expressions for$\cos (nx)$ and $\sin (nx)$ in terms of $\cos x$ and $\sin x$.

As written, the formula is not valid for non-integer powers $n$. However, there are generalizations of this formula valid for other exponents. These can be used to give explicit expressions for the ${{n}^{th}}$ roots of unity, that is, complex numbers$z$such that ${{z}^{n}}=1$.

So in question, we have to find the root of the complex number $5+12i$.
Let’s assume that the square root of any complex numbers is in the form $a+ib$.
So we get,
$a+ib=\sqrt{5+12i}$

Now squaring above to both sides we get,
${{\left( a+ib \right)}^{2}}=5+12i$
${{a}^{2}}+{{i}^{2}}{{b}^{2}}+2abi=5+12i$
${{a}^{2}}-{{b}^{2}}+2abi=5+12i$
So now comparing both sides we get,
${{a}^{2}}-{{b}^{2}}=5$ and $2ab=12$.
So ${{a}^{2}}-{{b}^{2}}=5$ and $ab=6$.
So solving the simultaneous equation to get the values of $a$ and$b$.
So substituting $b=\dfrac{6}{a}$ in above${{a}^{2}}-{{b}^{2}}=5$ we get,
${{a}^{2}}-{{\left( \dfrac{6}{a} \right)}^{2}}=5$

${{a}^{2}}-\left( \dfrac{36}{{{a}^{2}}} \right)=5$
${{a}^{4}}-36=5{{a}^{2}}$
${{a}^{4}}-5{{a}^{2}}-36=0$
So assuming the relation $x={{a}^{2}}$, we get the equation as :
${{x}^{2}}-5x-36=0$
So we know $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So here $a=1,b=-5,c=-36$.
$x=\dfrac{5\pm \sqrt{{{(-5)}^{2}}-4(1)(-36)}}{2(1)}=\dfrac{5\pm \sqrt{144}}{2(1)}=\dfrac{5\pm 12}{2(1)}$
So we get $x=\dfrac{5+12}{2},\dfrac{5-12}{2}$.
So$x=\dfrac{17}{2},\dfrac{-7}{2}$
So substituting in the original assumption, we get,
${{a}^{2}}=\dfrac{17}{2}$or${{a}^{2}}=-\dfrac{7}{2}$,
So we get $a=\sqrt{\dfrac{17}{2}}$or$a=\sqrt{-\dfrac{7}{2}}$
So here we can say that $a$ is real, since we assumed the root to be $a+ib$, where both, $a$ and $b$ were real numbers.
So we get $a=\sqrt{\dfrac{17}{2}}$.
Now substituting $a=\sqrt{\dfrac{17}{2}}$in${{a}^{2}}-{{b}^{2}}=5$.
So we get,
$\begin{align}
  & \dfrac{17}{2}-{{b}^{2}}=5 \\
 & \dfrac{17}{2}-5={{b}^{2}} \\
 & \dfrac{7}{2}={{b}^{2}} \\
\end{align}$
So we get $b=\sqrt{\dfrac{7}{2}}$.
So now $a+ib=\sqrt{\dfrac{17}{2}}+i\sqrt{\dfrac{7}{2}}$.
So $\sqrt{\dfrac{17}{2}}+i\sqrt{\dfrac{7}{2}}$ is the root of the complex number $5+12i$.

Note: Read the question carefully. So do not jumble yourself. Don’t make a mistake while substituting. You should be familiar with the core concept. So here $a$ is real and $b$ is imaginary. You should know to compare $a+ib=\sqrt{5+12i}$.
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