Question

square root of 3+4i complex number​

Answer 1

Answer:

Correct option is

C

2

(3−4i)

arg(3+4i)=tan

−1

(

3

4

)

New argument =

4

π

+tan

−1

(

3

4

)=tan

−1

(1)+tan

−1

(

3

4

)

=tan

−1

(

1−4/3

1+4/3

)=tan

−1

(

−1

4

)=tan

−1

(−7)

and if ∣z∣

2

=k

New ∣z∣

2

=2k

∣z∣=

2

k

New modulus is

2

times the old one. Therefore new complex number is

2

(1−7i)

Answer 2

Answer:

Here's your answer

Step-by-step explanation:

Let √3+4i

​=√x​+i√y​

Then (√3+4i​)2=(√x​+i√y​)2

⇒ 3 + 4i

=x − y + 2i√xy

Comparing real part and imaginary part, we get

3 = x − y and 2√xy = 4

⇒ xy = 4

∴ x + y = ±5     ( As x≠−5;x is real part )

∴ x = 4 , y = 1

⇒ √x ​ = ±2 , y = ±1

Hence square root is ±(2 + i)

Hope you understand.