Question

The square root of 5-12i is?​

Answer 1

Step-by-step explanation:

Suppose that a+bi is a square root of 5 + 12i. Then, (a+bi)^2 = (a^2 - b^2) + (2ab)i = 5 + 12i. 2ab = 12 ==> b = 6/a.

Answer 2

Answer:

The square root of the given complex number is

$\displaystyle{\boxed{\red{\sf\:\sqrt{5\:-\:12i}\:=\:\pm\:(\:\pm\:3\:\mp\:2\:i\:)}}}$

Step-by-step-explanation:

The given complex number is 5 - 12i.

We have to find the square root of this complex number.

The square root of a complex number a + bi is another complex number x + yi.

Let the square root of 5 - 12i be a complex number a + bi.

$\displaystyle{\sf\:\therefore\:\sqrt{5\:-\:12i}\:=\:a\:+\:bi}$

$\displaystyle{\implies\sf\:5\:-\:12i\:=\:(\:a\:+\:bi\:)^2\:\qquad\cdots[\:Squaring\:both\:sides\:]}$

$\displaystyle{\implies\sf\:5\:-\:12i\:=\:a^2\:+\:2ab\:i\:+\:(\:bi\:)^2}$

$\displaystyle{\implies\sf\:5\:-\:12i\:=\:a^2\:+\:2ab\:i\:+\:b^2\:i^2}$

$\displaystyle{\implies\sf\:5\:-\:12i\:=\:a^2\:+\:2ab\:i\:-\:b^2}$

$\displaystyle{\implies\sf\:5\:-\:12i\:=\:a^2\:-\:b^2\:+\:2ab\:i}$

Comparing both sides, we get,

$\displaystyle{\sf\:a^2\:-\:b^2\:=\:5}$

$\displaystyle{\implies\sf\:a^2\:=\:b^2\:+\:5\:\quad\:\cdots\:(\:1\:)}$

$\displaystyle{\sf\:2ab\:=\:-\:12}$

$\displaystyle{\implies\sf\:ab\:=\:-\:\cancel{\dfrac{12}{2}}}$

$\displaystyle{\implies\sf\:ab\:=\:-\:6}$

$\displaystyle{\implies\sf\:a\:=\:-\:\dfrac{6}{b}\:\quad\:\cdots\:(\:2\:)}$

By substituting this value in equation ( 1 ), we get,

$\displaystyle{\sf\:a^2\:=\:b^2\:+\:5}$

$\displaystyle{\implies\sf\:\left(\:-\:\dfrac{6}{b}\:\right)^2\:=\:b^2\:+\:5}$

$\displaystyle{\implies\sf\:\dfrac{36}{b^2}\:=\:b^2\:+\:5}$

$\displaystyle{\implies\sf\:\dfrac{36}{b^2}\:-\:b^2\:=\:5}$

$\displaystyle{\implies\sf\:\dfrac{36\:-\:b^4}{b^2}\:=\:5}$

$\displaystyle{\implies\sf\:36\:-\:b^4\:=\:5b^2}$

$\displaystyle{\implies\sf\:b^4\:+\:5b^2\:-\:36\:=\:0}$

$\displaystyle{\implies\sf\:b^4\:+\:9b^2\:-\:4b^2\:-\:36\:=\:0}$

$\displaystyle{\implies\sf\:b^2\:(\:b^2\:+\:9\:)\:-\:4\:(\:b^2\:+\:9\:)\:=\:0}$

$\displaystyle{\implies\sf\:(\:b^2\:+\:9\:)\:(\:b^2\:-\:4\:)\:=\:0}$

$\displaystyle{\implies\sf\:b^2\:+\:9\:=\:0\:\quad\:OR\:\quad\:b^2\:-\:4\:=\:0}$

$\displaystyle{\implies\sf\:b^2\:=\:-\:9\:\quad\:OR\:\quad\:b^2\:=\:4}$

$\displaystyle{\implies\boxed{\sf\:b\:=\:\pm\:3i}\sf\:\quad\:OR\:\quad\:\boxed{\sf\:b\:=\:\pm\:2}}$

But, b is a real number.

∴ b = ± 3i is unacceptable.

$\displaystyle{\therefore\:\underline{\boxed{\pink{\sf\:b\:=\:\pm\:2}}}}$

By substituting this value in equation ( 2 ), we get,

$\displaystyle{\sf\:a\:=\:-\:\dfrac{6}{b}}$

$\displaystyle{\implies\sf\:a\:=\:-\:\dfrac{\cancel{6}}{\pm\:\cancel{2}}}$

$\displaystyle{\implies\sf\:a\:=\:-\:(\:\pm\:3\:)}$

$\displaystyle{\implies\underline{\boxed{\blue{\sf\:a\:=\:\mp\:3}}}}$

∴ The square root of the given complex number is

$\displaystyle{\sf\:\sqrt{5\:-\:12i}\:=\:\pm\:(\:-\:3\:+\:2\:i\:)\:\quad\:OR\:\quad\:\sqrt{5\:-\:12i}\:=\:\pm\:(\:3\:-\:2\:i\:)}$

$\displaystyle{\underline{\boxed{\red{\sf\:\sqrt{5\:-\:12i}\:=\:\pm\:(\:\pm\:3\:\mp\:2\:i\:)}}}}$