Question

Class 11
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Math Complex Numbers
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1) Find The Square Root Of 4+3i
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Answer 1

♠hey there...

here is ur ans...⚡⚡

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Let the square root be a+bi. Then (a+bi)^2=(a^2-b^2)+2abi. So a^2-b^2=4 and 2ab=3.

Let a=3/(2b). Substitute into a^2-b^2=4 and simplify, yielding 4b^4+16b^2–9=0. Let c=b^2, which gives a quadratic in c. Solving the quadratic yields the solution c=-9/2, 1/2. Taking the square root yields b=3i/sqrt(2), 1/sqrt(2). Substitution yields a=-i/sqrt(2), 3/sqrt(2).

So there are two roots that when squared yield 4+3i — 3/sqrt(2)+i/sqrt(2) and -3/sqrt(2)-i/sqrt(2).


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hope it helps!!✌✌

Answer 2

Given : 4 + 3i

Now,

$= > \sqrt{4 + 3i} = x + iy$

On squaring both sides, we get

= > 4 + 3i = (x + iy)^2

We know that i^2 = -1.

= > 4 + 3i = x^2 - y^2 + 2ixy

= > 4 + 3i = (x^2 - y^2) + 2ixy

= > 4 = (x^2 - y^2) ------- (1)

and

2xy = 3 ----- (2)

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Now,

We know that (X^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2

= > (x^2 + y^2)^2 = (4)^2 + (3)^2

= > (x^2 + y^2)^2 = 16 + 9

= > (x^2 + y^2)^2 = 25

= > (x^2 + y^2) = 5 ------ (3)

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On solving (1) & (3), we get

= > x^2 - y^2 = 4

= > x^2 + y^2 = 5

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2x^2 = 9

x^2 = 9/2

$= > x = \frac{3}{\sqrt{2}}$

Substitute x in (3), we get

$= > (\frac{3}{\sqrt{2}})^2 + y^2 = 5$

$= > \frac{9}{2} + y^2 = 5$

$= > y^2 = 5 - \frac{9}{2}$

$= > y^2 = \frac{1}{2}$

$= > y = \sqrt{\frac{1}{2}}$

$= > y = \frac{1}{\sqrt{2}}$

$= > y = \frac{i}{\sqrt{2}} , -\frac{i}{\sqrt{2}}$

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Therefore,

$= > 4 + 3i = \boxed {\frac{3}{\sqrt{2}} + \frac{i}{\sqrt{2}} (or) {\frac{-3}{\sqrt{2}} - \frac{i}{\sqrt{2}}}}$

Hope this helps!