Question

Polar form of complex number 2+2i/1+√3i

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Answer 1

Answer:

√2 ( cos (-π/12) + i sin (-π/12) )

Step-by-step explanation:

Given : Complex number is (2+2i)/(1+√3i)

To find : Polar form of complex number

Solution :

Inorder to find the polar form of a complex number, firstly we have to calculate it's magnitude i.e. modulus and it's argument θ.

Let's simplify the complex number to easily calculate modulus and argument!

=> z = (2+2i)/(1+√3i)

=> z = (2+2i)/(1+√3i) × (1-√3i)/(1-√3i)

=> z = [2(1-√3i)+2i(1-√3i)]/[(1+√3i)(1-√3i)]

=> z = (2-2√3i+2i-2√3i²)/(1+√3²)

=> z = (2-2√3i+2i+2√3)/(1+3) ( i² = -1 )

=> z = 2(1-√3i+i+√3)/2(2)

=> z = [1+√3 -(√3-1)i]/2

=> z = (1+√3)/2 - (√3-1)i/2

Let's assume that, x = real part of complex number = (1+√3)/2 and y = imaginary part of complex number = -(√3-1)/2. (Note that the given complex number lies in the 2nd quadrant as it's x coordinate is +ve and y coordinate is negative.)

Modulus of complex number is given by,

=> |z| = √{(x² + y²)}

=> |z| = √{[(1+√3)/2]² + [-(√3-1)/2]²}

=> |z| = √{(1+3+2√3)/4 + (√3²+1-2√3)/4}

=> |z| = √{(4+2√3)/4 + (3+1-2√3)/4}

=> |z| = √{(4+2√3+4-2√3)/4}

=> |z| = √{8/4}

=> |z| = √{2}

=> |z| = √2

Argument of complex number in 4th quadrant is given by,

=> Arg(z) = -θ , Here θ = arctan |y/x| ( arctan is tan inverse )

=> Arg(z) = - arctan | [-(√3-1)/2]/[(√3+1)/2] |

=> Arg(z) = - arctan (√3-1)/(√3+1)

=> Arg(z) = - arctan (√3-1)/(√3+1) × (√3-1)/(√3-1)

=> Arg(z) = - arctan ( 3+1-2√3)/(3-1)

=> Arg(z) = - arctan ( 4-2√3)/2

=> Arg(z) = - arctan ( 2 - √3)

=> Arg(z) = - π/12

The polar form of a complex number is given by,

|z| ( cos θ + i sin θ ) where θ is argument and |z| is magnitude.

=> √2 ( cos (-π/12) + i sin (-π/12) )

This is the required polar form.