Question

If $(a+ib)=\frac{1+i}{1-i}$, then prove that (a² + b²)=1.

Answer 1

Answer:

Step-by-step explanation:

Hi,

Given that a + ib = (1+i)/(1 - i)----(1)

To get a conjugate  of complex number simply, we need to

replace i in every term by -i

So, Conjugate of a + ib is a - ib,

a - ib = (1-i)/(1 + i)------(2)

Multiplying (1) and (2), we get

(a + ib)*(a - ib) = (1+i)/(1 - i)*(1-i)/(1 + i)

= 1

(a² - (ib)²) = 1

a² + b² = 1.

Hence, Proved that a² + b² = 1

Hope, it helps !

Answer 2

Answer:

Step-by-step explanation:

Concept:

Modulus of a complex number a+ib is deflined as

$|a+ib| = \sqrt{a^2+b^2}$

If $z_1\:and\:z_2$ are two complex numbers

then

$|\frac{z_1}{z_2}|=\frac{|z_1|}{|z_2|}$

Given:

$a+ib=\frac{1+i}{1-i}$

Take modulus on both sides

$|a+ib|=|\frac{1+i}{1-i}|\\\\|a+ib|=\frac{|1+i|}{|1-i|}\\\\\sqrt{a^2+b^2}=\frac{\sqrt{1+1}}{\sqrt{1+1}}\\\\\sqrt{a^2+b^2}=1$

squaring on both sides

$a^2+b^2=1$