第11类RD Sharma解决方案–第13章复数–练习13.3

问题1.找到以下复数的平方根。

(i)– 5 + 12i

(ii)-7 – 24i

(iii)1 –我

(iv)– 8 – 6i

(v)8 – 15i

(六) $-11 - 60\sqrt-1$

(vii) $1 + 4\sqrt3$

(viii)4i

(ix)-i

解决方案：

如果b> 0， $\sqrt{a+ib}=\pm\left[\left(\frac{a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}\right]$

如果b <0， $\sqrt{a+ib}=\pm\left[\left(\frac{a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}\right]$

(i)– 5 + 12i

Given:

– 5 + 12i

We know, Z = a + ib

So, $\sqrt{a + ib} = \sqrt{-5+12i}$

Here, b > 0

Let us simplify now,

$\sqrt{-5+12i}=\pm\left[\left(\frac{-5+\sqrt{(-5)^2+12^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{(-5)^2+12^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{-5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+\sqrt{169}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{169}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+13}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+13}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{18}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i9^{\frac{1}{2}}\right]\\ =\pm[2+3i]$

∴ Square root of (– 5 + 12i) is ±[2 + 3i]

(ii)-7 – 24i

Given:

-7 – 24i

We know, Z = -7 – 24i

So, $\sqrt{a + ib} = \sqrt{-7 - 24i}$

Here, b < 0

Let us simplify now,

$\sqrt{-7-24i}=\pm\left[\left(\frac{-7+\sqrt{(-7)^2+(-24)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{(-7)^2+(-24)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-7+\sqrt{625}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{625}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-7+25}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+25}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{18}{2}\right)^{\frac{1}{2}}-i\left(\frac{32}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[9^{\frac{1}{2}}-i16^{\frac{1}{2}}\right]\\ =\pm[3-4i]$

∴ Square root of (-7 – 24i) is ± [3 – 4i]

(iii)1 –我

Given:

1 – i

We know, Z = (1 – i)

So, $\sqrt{a + ib} = \sqrt{1 - i}$

Here, b < 0

Let us simplify now,

$\sqrt{1-i}=\pm\left[\left(\frac{1+\sqrt{(1)^2+(-1)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{(1)^2+(-1)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+\sqrt{2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{2}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\sqrt{\frac{\sqrt2+1}{2}}\right)-i\left(\sqrt{\frac{\sqrt2-1}{2}}\right)\right]$

∴ Square root of (1 – i) is ± $\pm\left[\left(\sqrt{\frac{\sqrt2+1}{2}}\right)-i\left(\sqrt{\frac{\sqrt2-1}{2}}\right)\right]$

(iv)-8 -6i

Given:

-8 -6i

We know, Z = -8 -6i

So, $\sqrt{a + ib}$ = -8 -6i

Here, b < 0

Let us simplify now,

$\sqrt{-8-6i}=\pm\left[\left(\frac{-8+\sqrt{(-8)^2+(-6)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{(-8)^2+(-6)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-8+\sqrt{100}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{100}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-8+10}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+10}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{2}{2}\right)^{\frac{1}{2}}-i\left(\frac{10}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[1^{\frac{1}{2}}-i9^{\frac{1}{2}}\right]\\ =\pm[1-3i]$

∴ Square root of (-8 -6i) is ± [1 – 3i]

(v)8 – 15i

Given:

8 – 15i

We know, Z = 8 – 15i

So, $\sqrt{a + ib}$  = 8 – 15i

Here, b < 0

Let us simplify now,

$\sqrt{8-15i}=\pm\left[\left(\frac{8+\sqrt{(8)^2+(-15)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{(8)^2+(-15)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{8+\sqrt{64+225}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{64+225}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8+\sqrt{289}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{289}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8+17}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+17}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{25}{2}\right)^{\frac{1}{2}}-i\left(\frac{9}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\frac{5}{\sqrt2}-\frac{13}{\sqrt2}\right]\\ =\pm\frac{1}{\sqrt2}(5-3i)$

∴ Square root of (8 – 15i) is ± $\frac{1}{\sqrt2}(5-3i)$

(六) $-11 - 60\sqrt{-1}$

Given:

$-11 - 60\sqrt{-1}$

We know, Z = $-11 - 60\sqrt{-1}$

So, $\sqrt{a + ib} = -11 - 60\sqrt{-1}$

= -11 – 60i

Here, b < 0

Let us simplify now,

$\sqrt{-11-60i}=\pm\left[\left(\frac{-11+\sqrt{(-11)^2+(-60)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{(-11)^2+(60)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-11+\sqrt{121+3600}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{121+3600}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-11+\sqrt{3721}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{3721}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-11+61}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+61}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{50}{2}\right)^{\frac{1}{2}}-i\left(\frac{72}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[25^{\frac{1}{2}-i36^{\frac{1}{2}}}\right]\\ =\pm(5-6i)$

∴ Square root of ($-11 - 60\sqrt{-1}$) is ± (5 – 6i)

(vii) $1 + 4\sqrt{-3}$

Given:

$1 + 4\sqrt{-3}$

We know, Z = $1 + 4\sqrt{-3}$

So, $\sqrt{a + ib} = 1 + 4\sqrt{-3}$

$= 1 + 4(\sqrt{3}) (\sqrt{-1})\\ = 1 + 4\sqrt{3i}$

Here, b > 0

Let us simplify now,

$\sqrt{1+4\sqrt3i}=\pm\left[\left(\frac{1+\sqrt{(1)^2+(4\sqrt3)^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{1^2+(4\sqrt3)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+\sqrt{49}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{49}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+7}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+7}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{6}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i3^{\frac{1}{2}}\right]\\ =\pm[2+\sqrt3i]$

∴ Square root of $(1 + 4\sqrt{-3})$ is ± $(2+\sqrt3i)$

(viii)4i

Given:

4i

We know, Z = 4i

So, $\sqrt{a + ib}$ = 4i

Here, b > 0

Let us simplify now,

$\sqrt{4i}=\pm\left[\left(\frac{0+\sqrt{(0)^2+(4)^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{0^2+(4)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+4}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{4}{2}\right)^{\frac{1}{2}}+i\left(\frac{4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[2^{\frac{1}{2}}+i2^{\frac{1}{2}}\right]\\ =\pm[\sqrt2+\sqrt2i]\\ =\pm\sqrt2(1+i)$

∴ Square root of 4i is ± $\sqrt{2 (1 + i)}$

(ix)–i

Given:

-i

We know, Z = -i

So, $\sqrt{(a + ib)}$ = -i

Here, b < 0

Let us simplify now,

$\sqrt{-i}=\pm\left[\left(\frac{0+\sqrt{0^2+(-1)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0^2+(-1)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+1}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+1}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1}{2}\right)^{\frac{1}{2}}-i\left(\frac{1}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ =\pm\frac{1}{\sqrt2}(1-i)$

∴ Square root of –i is ± $\frac{1}{\sqrt2} (1 - i)$