Question

Find the modulus and argument of the complex number $\frac{1+2i}{1-2i}$.

Answer 1

Answer:

Modulus = 1

Amplitude = -37°

Step-by-step explanation:

Hi,

If z = a + ib is any complex number such that a ∈R, b∈R  

and i = √-1,

then the modulus of complex number, |Z| is given by

|Z| = √a² + b²

Amplitude of the complex number , θ is given by,

θ = tan⁻¹(b/a)

Consider (1 + 2i)/(1 - 2i)

Multiplying and dividing by 1 + 2i, we get

(1 + 2i)*(1 + 2i)/(1 - 2i)*(1 + 2i)

= (1 + 4i -4)/(1 + 4)

= -3/5 + i(4/5)

Modulus of z, |Z| = √(-3/5)² + (4/5)²

= √(9/25) + (16/25)

= √25/25

= 1

Amplitude, ∅ = tan⁻¹((4/5)/(-3/5))

= tan⁻¹(-3/4)

= -tan⁻¹(3/4)

= -37°

Hope, it helps !

Answer 2

Answer:

Step-by-step explanation:

Concept:

The modulus of a complex number

a+ib is defined by

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

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

The argument of z=a+ib is defined by

$argz=tan^{-1}\frac{b}{a}$

$arg (z_1z_2)=argz_1-argz_2$

$z=\frac{1+2i}{1-2i}\\\\|z|=|\frac{1+2i}{1-2i}|\\\\|z|=\frac{|1+2i|}{|1-2i|}\\\\|z|=\frac{\sqrt{1+4}}{\sqrt{1+4}}$

|z|= 1

$arg(\frac{1+2i}{1-2i})=arg(1+2i)-arg(1-2i)\\\\=tan^{-1}(\frac{2}{1})-tan^{-1}(\frac{2}{-1})\\\\=tan^{-1}(2)-tan^{-1}(-2)\\\\=tan^{-1}(2)-tan^{-1}(-2)\\\\=tan^{-1}(2)+tan^{-1}(2)\\\\=2tan^{-1}(2)$