Question

Find the square root of (- 20 – 20i).

Answer 1

Hint: Take $\sqrt{\left( -21-20i \right)}=a+bi$. Squaring and simplifying, find the real and imaginary part. Thus find the value of a and b by solving the equation formed. Substitute the values in (a + bi) to get the square root.
Complete step by step answer:
Every complex number has a complex square root. We have been asked to find the square root of the complex number (-21 – 20i). We know that all square roots of the number will satisfy the equation $-21-20i={{x}^{2}}$ by definition of a square root.
We also know that x can be expressed as (a + bi), where a and b are real numbers. Since, the square roots of a complex number are always complex.
So let us take $\sqrt{\left( -21-20i \right)}=\left( a+bi \right)$
Let us square on both sides.
$-21-20i={{\left( a+bi \right)}^{2}}$
We know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\Rightarrow -21-20i={{a}^{2}}+2abi+{{\left( ib \right)}^{2}}$ ($\because {{i}^{2}}=-1$)
$\Rightarrow -21-20i={{a}^{2}}-{{b}^{2}}+2abi$
Now both sides of the equation are the same. Let us compare the real part of the equation. Thus we get,
${{a}^{2}}-{{b}^{2}}=-21-(1)$
Now let us compare the imaginary part of the equation. We get,

& 2abi=-20i\Rightarrow 2ab=-20 \\\ & \Rightarrow ab=-10-(2) \\\ \end{aligned}$$ We now have two equations with 2 unknowns. We can solve these simultaneous equations for a and b. From (2) we can say that, $$b=\dfrac{-10}{a}$$. Now let us substitute this in equation (1). $$\begin{aligned} & {{a}^{2}}-{{b}^{2}}=-21 \\\ & \Rightarrow {{a}^{2}}-{{\left( \dfrac{-10}{a} \right)}^{2}}=-21 \\\ & \Rightarrow {{a}^{2}}-\dfrac{100}{{{a}^{2}}}=-21 \\\ & \Rightarrow {{a}^{4}}-100=-21{{a}^{2}} \\\ & \Rightarrow {{a}^{4}}+21{{a}^{2}}-100=0 \\\ \end{aligned}$$ We can split the second term $$21{{a}^{2}}$$ as $$\left( 25-4 \right){{a}^{2}}$$. $$\begin{aligned} & \therefore {{a}^{2}}+\left( 25-4 \right){{a}^{2}}-100=0 \\\ & {{a}^{2}}+25{{a}^{2}}-4{{a}^{2}}-100=0 \\\ & {{a}^{2}}\left( {{a}^{2}}+25 \right)-4\left( {{a}^{2}}+25 \right)=0 \\\ & \Rightarrow \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}+25 \right)=0 \\\ \end{aligned}$$ Thus $${{a}^{2}}-4=0$$ $$\begin{aligned} & \Rightarrow {{a}^{2}}=4 \\\ & \therefore a=\sqrt{4}=\pm 2 \\\ \end{aligned}$$ And $${{a}^{2}}+25=0$$ $${{a}^{2}}=-25$$ We assumed a to be a real number. So, $${{a}^{2}}=-25$$ has no solutions. Thus we got a as 2 and -2. $$\therefore $$ a = 2 and a = -2. Now let us substitute the value of a in equation (2). $$\begin{aligned} & ab=-10 \\\ & \Rightarrow b=\dfrac{-10}{a} \\\ \end{aligned}$$ When a = 2, $$b=\dfrac{-10}{2}=-5$$ a = -2, $$b=\dfrac{-10}{\left( -2 \right)}=5$$ Thus by putting a and b in (a + bi), we get the square roots of (-21 - 20i) as (2 - 5i) and (-2 + 5i) $$=\pm \left( 2-5i \right)$$. $$\therefore $$ The square root is (-21 – 20i) = (2 – 5i) and (-2 + 5i). Note: We can also solve it as, We got real part as, $${{a}^{2}}-{{b}^{2}}=-21-(1)$$ Imaginary part as, $$2xy=-20-(2)$$ $$\begin{aligned} & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}+4{{a}^{2}}{{b}^{2}} \\\ & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( -21 \right)}^{2}}+{{\left( -20 \right)}^{2}}=841 \\\ & \therefore {{a}^{2}}+{{b}^{2}}=\sqrt{841}=29-(3) \\\ \end{aligned}$$ Adding (1) and (3), we get $$\begin{aligned} & 2{{a}^{2}}=8 \\\ & \therefore a=\pm 2 \\\ \end{aligned}$$ Thus we get, $$b=\pm 5$$. $$\therefore $$ Square root of -21 – 20i = $$\pm \left( 2-5i \right)$$.