Question

Please sove this ques​

Answer 1

$\boxed{\begin {minipage}{1.5cm} i^2=-1 \\\\ i^3=-i \\\\ i^4=1 \\\\ i^5=i \end{minipage}}$

Thus

$6i^5+7i^4+3i^3+5i^2-4=6i+7(1)+3(-i)+5(-1)-4\\\\6i^5+7i^4+3i^3+5i^2-4=6i+7-3i-5-4\\\\6i^5+7i^4+3i^3+5i^2-4=\mathbf{-2+3i}$

Now we have it in the form $a+ib.$

Let's find its square root.

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Let us have,

$x+iy=\sqrt{a+ib}$

where $a$ and $b$ are known and $x$ and $y$ have to be found. Then,

$(x+iy)^2=a+ib\\\\x^2-y^2+2xyi=a+ib$

Equating like coefficients,

$a=x^2-y^2\quad\longrightarrow\quad (1)\\\\b=2xy\quad\longrightarrow\quad (2)$

Then,

$a^2=(x^2-y^2)^2\\\\a^2=x^4+y^4-2x^2y^2\\\\a^2+b^2=x^4+y^4-2x^2y^2+4x^2y^2\quad [\text {From (2)}]\\\\a^2+b^2=x^4+y^4+2x^2y^2\\\\a^2+b^2=(x^2+y^2)^2\\\\\sqrt{a^2+b^2}=x^2+y^2\quad\longrightarrow\quad (3)$

(3) shows that the modulus of square of a complex number is the square of its modulus.

Adding (1) and (3), we get,

$x=\pm\sqrt{\dfrac {\sqrt{a^2+b^2}+a}{2}}\\\\\\y=\pm\sqrt{\dfrac {\sqrt{a^2+b^2}-a}{2}}$

Let $\sqrt{a^2+b^2}=r.$ Then,

$x=\pm\sqrt{\dfrac {r+a}{2}}\\\\\\y=\pm\sqrt{\dfrac {r-a}{2}}$

But we should note the signs of real and imaginary parts too. Then, we get that,

$\boxed{\sqrt{a+ib}=\left\{\begin{array}{l}\pm\left(\sqrt{\dfrac {r+a}{2}}+i\sqrt{\dfrac {r-a}{2}}\right),\ b\ \textgreater\ 0\\\\\pm\left(\sqrt{\dfrac {r+a}{2}}-i\sqrt{\dfrac {r-a}{2}}\right),\ b\ \textless\ 0\end{array}\right.}$

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Here,

$a=-2\ \textless\ 0\\\\b=3\ \textgreater\ 0$

$r=\sqrt{(-2)^2+3^2}=\sqrt{4+9}=\sqrt{13}$

Since $b\ \textgreater\ 0,$

$\sqrt{a+ib}=\pm\left(\sqrt{\dfrac {r+a}{2}}+i\sqrt{\dfrac {r-a}{2}}\right)\\\\\\\boxed{\sqrt{-2+3i}=\pm\left(\sqrt{\dfrac {\sqrt{13}-2}{2}}+i\sqrt{\dfrac {\sqrt{13}+2}{2}}\right)}}$

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