If a-i b-x-i2-2 x2-1 then prove that a2-b2-x2-12-2 x2-12

We have (a+ib)=(x+i)^2/(2x^2+1) =(x^2+i^2+2ix)/(2x^2+1) =(x^2 1)+i(2x)/(2x^2+1) ⇒ (a+ib)=(x^2 1)/(2x^2 1)+i .2x/(2x^2+1) ⇒ |a+ib|^2= | (x^2 1)/(2x^2+1)+i. 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Iota

If a+i b=x+i2...

Question

If $(a + i b) = \frac{(x + i)^{2}}{(2 x^{2} + 1)}$ then prove that $(a^{2} + b^{2}) = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$ .

Open in App

Solution

We have

$(a + i b) = \frac{(x + i)^{2}}{(2 x^{2} + 1)} = \frac{(x^{2} + i^{2} + 2 i x)}{(2 x^{2} + 1)} = \frac{(x^{2} - 1) + i (2 x)}{(2 x^{2} + 1)}$

$\Rightarrow (a + i b) = \frac{(x^{2} - 1)}{(2 x^{2} - 1)} + i . \frac{2 x}{(2 x^{2} + 1)}$

$\Rightarrow | a + i b |^{2} = {∣ ∣ ∣ \frac{(x^{2} - 1)}{(2 x^{2} + 1)} + i . \frac{2 x}{(2 x^{2} + 1)} ∣ ∣ ∣}^{2}$

$\Rightarrow (a^{2} + b^{2}) = {\frac{(x^{2} - 1)^{2}}{(2 x^{2} + 1)^{2}} + \frac{4 x^{2}}{(2 x^{2} + 1)^{2}}} = \frac{(x^{2} - 1)^{2} + 4 x^{2}}{(2 x^{2} + 1)^{2}} = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$

Hence, $(a^{2} + b^{2}) = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$ .

Suggest Corrections

Similar questions

Q. If

a + i b = \frac{{(x + i)}^{2}}{{2 x}^{2} + 1}

, prove that

a^{2} + b^{2} = \frac{{(x^{2} + 1)}^{2}}{{(2 x^{2} + 1)}^{2}}

Q. If

x + i y = \sqrt{\frac{a + i b}{c + i d}}

, prove that

(x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}

Q. Let

(x + i y) = \sqrt{\frac{a + i b}{c + i d}}

, then prove that

(x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}

Q. If

(a + i b) = \frac{1 + i}{1 - i}

, then prove that

(a^{2} + b^{2}) = 1

Q. If

a + i b = \frac{p + i}{2 q - i}

then prove that

a^{2} + b^{2} = \frac{p^{2} + 1}{4 q^{2} + 1}

Join BYJU'S Learning Program

If (a+ib)=(x+i)2(2x2+1) then prove that (a2+b2)=(x2+1)2(2x2+1)2.

We have (a+ib)=(x+i)2(2x2+1)=(x2+i2+2ix)(2x2+1)=(x2−1)+i(2x)(2x2+1) ⇒(a+ib)=(x2−1)(2x2−1)+i.2x(2x2+1) ⇒|a+ib|2=∣∣∣(x2−1)(2x2+1)+i.2x(2x2+1)∣∣∣2 ⇒(a2+b2)={(x2−1)2(2x2+1)2+4x2(2x2+1)2}=(x2−1)2+4x2(2x2+1)2=(x2+1)2(2x2+1)2 Hence, (a2+b2)=(x2+1)2(2x2+1)2.

If $(a + i b) = \frac{(x + i)^{2}}{(2 x^{2} + 1)}$ then prove that $(a^{2} + b^{2}) = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$ .