Question

Find the square root of the following complex numbers

i)7 + 24i \\\ ii)1 + 4\sqrt 3 i \\\ iii)3 + 2\sqrt {10} i $$

Answer 1

For finding the square root of a complex number we will assume the square roots of the number in the form of two variables in the rectangular system to be$a + bi$. Later on we can compare it with original numbers in order to calculate the values of $a$and $b$ which will give the square root. It can be done by squaring both sides and then equating the real part with the real number and the imaginary part with the imaginary number.

Formula used:
As we know that the complex numbers are represented as $a + bi$ so we will expand the equation in this format and then substitute the value of${i^2} = - 1$. Later on by simplifying the equation by squaring and equating both sides we will get the result.

Complete step by step solution:
$i)7 + 24i$
Let $\sqrt {7 + 24i} = a + bi{\text{ where a,b}} \in {\text{R}}$
Now after squaring both the sides we will get
$7 + 24i = {(a + bi)^2}$
Therefore we get

\Rightarrow 7 + 24i = ({a^2} - {b^2}) + 2abi \\\ $$ Now $${i^2} = - 1$$ Later on we will equate real and imaginary parts and we will get $${a^2} - {b^2} = 7$$ and $$2ab = 24$$ Therefore $${a^2} - {b^2} = 7$$ and $$b = \dfrac{{12}}{a}$$ $${a^2} - {\left( {\dfrac{{12}}{a}} \right)^2} = 7 \\\ \Rightarrow {a^2} - \dfrac{{144}}{{{a^2}}} = 7 \\\ \Rightarrow {a^4} - 144 = 7{a^2} \\\ \Rightarrow {a^2} - 7{a^2} - 144 = 0 \\\ \Rightarrow ({a^2} - 16)({a^2} + 9) = 0 \\\ \Rightarrow {a^2} = 16{\text{ or }}{a^2} = - 9 \\\ $$ But as we know $$a \in R$$ which means $${a^2} \ne - 9$$ $${a^2} = 16 \\\ \Rightarrow a = \pm 4 $$ Now putting values of $$a$$ we get If $$a = 4$$ then $$b = 3$$ If $$a = - 4$$ then $$b = - 3$$ **So the square root of $$7 + 24i$$ will be $$ \pm (4 + 3i)$$.** $$ii)1 + 4\sqrt 3 i$$ Let $$\sqrt {1 + 4\sqrt 3 i} = a + bi{\text{ where a,b}} \in {\text{R}}$$ Now after squaring both the sides we will get $$1 + 4\sqrt 3 i = {(a + bi)^2} \\\ \Rightarrow 1 + 4\sqrt 3 i = {a^2} + {b^2}{i^2} + 2abi \\\ \Rightarrow 1 + 4\sqrt 3 i - ({a^2} - {b^2}) + 2abi \\\ $$ Now $${i^2} = - 1$$ Later on we will equate real and imaginary parts and we will get $${a^2} - {b^2} = 1$$and $$2ab = 4\sqrt 3 $$ Therefore $${a^2} - {b^2} = 1$$and $$b = \dfrac{{2\sqrt 3 }}{a}$$ $${a^2} - {\left( {\dfrac{{2\sqrt 3 }}{a}} \right)^2} = 1 \\\ \Rightarrow {a^2} - \dfrac{{12}}{{{a^2}}} = 1 \\\ \Rightarrow {a^4} - 12 = {a^2} \\\ \Rightarrow {a^4} - {a^2} - 12 = 0 \\\ \Rightarrow ({a^2} - 4)({a^2} + 3) = 0 \\\ \Rightarrow {a^2} = 4{\text{ or }}{a^2} = - 3 \\\ $$ But as we know $$a \in R$$ which means $${a^2} \ne - 3$$ $${a^2} = 4 \\\ \Rightarrow a = \pm 2 $$ Now putting values of $$a$$ we get If $$a = 2$$ then $$b = \sqrt 3 $$ If $$a = - 2$$ then $$b = - \sqrt 3 $$ **So the square root of $$1 + 4\sqrt 3 i$$ will be $$ \pm (2 + \sqrt 3 i)$$.** $$iii)3 + 2\sqrt {10} i$$ Let $$\sqrt {3 + 2\sqrt {10} i} = a + bi{\text{ where a,b}} \in {\text{R}}$$ Now after squaring both the sides we will get $$3 + 2\sqrt {10} i = {(a + bi)^2}$$ Therefore we get $$3 + 2\sqrt {10} i = {a^2} + {b^2}{i^2} + 2abi \\\ \Rightarrow 3 + 2\sqrt {10} i = ({a^2} - {b^2}) + 2abi \\\ $$ Now $${i^2} = - 1$$ Later on we will equate real and imaginary parts and we will get $${a^2} - {b^2} = 3$$and $$2ab = 2\sqrt {10} $$ Therefore $${a^2} - {b^2} = 3$$and $$b = \dfrac{{\sqrt {10} }}{a}$$ $${a^2} - {\left( {\dfrac{{\sqrt {10} }}{a}} \right)^2} = 3 \\\ \Rightarrow {a^2} - \dfrac{{10}}{{{a^2}}} = 3 \\\ \Rightarrow {a^4} - 10 = 3{a^2} \\\ \Rightarrow {a^2} - 3{a^2} - 10 = 0 \\\ \Rightarrow {a^2} = 5{\text{ or }}{a^2} = - 5 \\\ $$ But as we know $$a \in R$$ which means $${a^2} \ne - 5$$ $${a^2} = 5 \\\ \Rightarrow a = \pm \sqrt 5 $$ Now putting values of $$a$$ we get If $$a = \sqrt 5 $$ then $$b = \sqrt 2 $$ If $$a = - \sqrt 5 $$ then $$b = - \sqrt 2 $$ **So the square root of $$3 + 2\sqrt {10} i$$ will be $$ \pm (\sqrt 5 + \sqrt 2 i)$$.** **Note:** Here while solving the above equation we should keep in mind that in a complex number $$a + bi$$, $$a$$ and $$b$$ are the real numbers and hence we never consider value of $$a$$ as negative like $$ - 3$$ . We should try to avoid the calculation mistakes. Keep in mind that here the value of $${i^2} = - 1$$so while solving the equation $${(a + bi)^2}$$ replace the value of$${i^2}$$ as $$ - 1$$.