Answer
Verified
426.3k+ views
Hint: We start solving the problem by assuming the square root of the given complex number. We then square them on both sides and compare the real and imaginary parts on both sides. We then make the necessary substitutions and calculations related to the real and imaginary parts of the square root to get the required result.
Complete step-by-step solution:
According to the problem, we need to find the square root of the complex number \[-5+12i\].
Let us assume the square root of the given complex number \[-5+12i\] be \[a+bi\].
So, we get \[a+bi=\sqrt{-5+12i}\] ---(1).
Let us do square on both sides.
$\Rightarrow {{\left( a+bi \right)}^{2}}=-5+12i$.
$\Rightarrow {{a}^{2}}+{{\left( ib \right)}^{2}}+2\left( a \right)\left( ib \right)=-5+12i$.
$\Rightarrow {{a}^{2}}+{{i}^{2}}{{b}^{2}}+2iab=-5+12i$.
We know that ${{i}^{2}}=-1$.
$\Rightarrow {{a}^{2}}+\left( -1 \right){{b}^{2}}+2iab=-5+12i$.
$\Rightarrow {{a}^{2}}-{{b}^{2}}+2iab=-5+12i$.
$\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)+\left( 2ab \right)i=-5+12i$.
Comparing real and imaginary parts on both sides, we get ${{a}^{2}}-{{b}^{2}}=-5$ and $2ab=12$.
Now, we have $2ab=12\Leftrightarrow b=\dfrac{6}{a}$ ---(2). Let us substitute this result in ${{a}^{2}}-{{b}^{2}}=-5$.
So, we get ${{a}^{2}}-{{\left( \dfrac{6}{a} \right)}^{2}}=-5$.
$\Rightarrow {{a}^{2}}-\dfrac{36}{{{a}^{2}}}=-5$.
$\Rightarrow \dfrac{{{a}^{4}}-36}{{{a}^{2}}}=-5$.
$\Rightarrow {{a}^{4}}-36=-5{{a}^{2}}$.
$\Rightarrow {{a}^{4}}+5{{a}^{2}}-36=0$.
\[\Rightarrow {{a}^{4}}+9{{a}^{2}}-4{{a}^{2}}-36=0\].
\[\Rightarrow {{a}^{2}}\left( {{a}^{2}}+9 \right)-4\left( {{a}^{2}}+9 \right)=0\].
\[\Rightarrow \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}+9 \right)=0\].
\[\Rightarrow {{a}^{2}}-4=0\] or \[{{a}^{2}}+9=0\].
\[\Rightarrow {{a}^{2}}=4\] or \[{{a}^{2}}=-9\].
\[\Rightarrow a=\pm 2\] (we know that ${{a}^{2}}$ is always greater than zero so, we neglect \[{{a}^{2}}=-9\]).
Now, let us find the value of b by substituting the value of “a” in equation (2).
So, we get $b=\dfrac{6}{\pm 2}\Leftrightarrow b=\pm 3$.
Let us substitute this in equation (1) to get the square roots of the given complex number.
So, we get \[\pm \left( 2+3i \right)=\sqrt{-5+12i}\].
So, we have found the square root of the given complex number \[-5+12i\] as \[\pm \left( 2+3i \right)\].
$\therefore$ The square root of the complex number \[-5+12i\] is \[\pm \left( 2+3i \right)\].
Note: We should note that in a complex number $a+ib$, ‘a’ and ‘b’ are real numbers and this is the reason why we didn’t consider \[{{a}^{2}}=-9\]. We should not make calculation mistakes while solving this problem. We can also solve the problem by finding the exponential form of the given complex number and then find the square root of that exponential form which will require a good amount of calculation. Similarly, we can expect problems to find the Euler and exponential form of the given complex number.
Complete step-by-step solution:
According to the problem, we need to find the square root of the complex number \[-5+12i\].
Let us assume the square root of the given complex number \[-5+12i\] be \[a+bi\].
So, we get \[a+bi=\sqrt{-5+12i}\] ---(1).
Let us do square on both sides.
$\Rightarrow {{\left( a+bi \right)}^{2}}=-5+12i$.
$\Rightarrow {{a}^{2}}+{{\left( ib \right)}^{2}}+2\left( a \right)\left( ib \right)=-5+12i$.
$\Rightarrow {{a}^{2}}+{{i}^{2}}{{b}^{2}}+2iab=-5+12i$.
We know that ${{i}^{2}}=-1$.
$\Rightarrow {{a}^{2}}+\left( -1 \right){{b}^{2}}+2iab=-5+12i$.
$\Rightarrow {{a}^{2}}-{{b}^{2}}+2iab=-5+12i$.
$\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)+\left( 2ab \right)i=-5+12i$.
Comparing real and imaginary parts on both sides, we get ${{a}^{2}}-{{b}^{2}}=-5$ and $2ab=12$.
Now, we have $2ab=12\Leftrightarrow b=\dfrac{6}{a}$ ---(2). Let us substitute this result in ${{a}^{2}}-{{b}^{2}}=-5$.
So, we get ${{a}^{2}}-{{\left( \dfrac{6}{a} \right)}^{2}}=-5$.
$\Rightarrow {{a}^{2}}-\dfrac{36}{{{a}^{2}}}=-5$.
$\Rightarrow \dfrac{{{a}^{4}}-36}{{{a}^{2}}}=-5$.
$\Rightarrow {{a}^{4}}-36=-5{{a}^{2}}$.
$\Rightarrow {{a}^{4}}+5{{a}^{2}}-36=0$.
\[\Rightarrow {{a}^{4}}+9{{a}^{2}}-4{{a}^{2}}-36=0\].
\[\Rightarrow {{a}^{2}}\left( {{a}^{2}}+9 \right)-4\left( {{a}^{2}}+9 \right)=0\].
\[\Rightarrow \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}+9 \right)=0\].
\[\Rightarrow {{a}^{2}}-4=0\] or \[{{a}^{2}}+9=0\].
\[\Rightarrow {{a}^{2}}=4\] or \[{{a}^{2}}=-9\].
\[\Rightarrow a=\pm 2\] (we know that ${{a}^{2}}$ is always greater than zero so, we neglect \[{{a}^{2}}=-9\]).
Now, let us find the value of b by substituting the value of “a” in equation (2).
So, we get $b=\dfrac{6}{\pm 2}\Leftrightarrow b=\pm 3$.
Let us substitute this in equation (1) to get the square roots of the given complex number.
So, we get \[\pm \left( 2+3i \right)=\sqrt{-5+12i}\].
So, we have found the square root of the given complex number \[-5+12i\] as \[\pm \left( 2+3i \right)\].
$\therefore$ The square root of the complex number \[-5+12i\] is \[\pm \left( 2+3i \right)\].
Note: We should note that in a complex number $a+ib$, ‘a’ and ‘b’ are real numbers and this is the reason why we didn’t consider \[{{a}^{2}}=-9\]. We should not make calculation mistakes while solving this problem. We can also solve the problem by finding the exponential form of the given complex number and then find the square root of that exponential form which will require a good amount of calculation. Similarly, we can expect problems to find the Euler and exponential form of the given complex number.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
10 examples of evaporation in daily life with explanations
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE