Question #2824

Square root of 11 + 5i in polar form with r greater than or equal to 0 and theta in the interval 0 less than or equal to theta less than or equal to 2 pie and rounded to the nearest thousandth.

Expert's answer

Suppose

√(11+5i)=r(cos +isinφ)

We have that

11+5i=A(cosφ+isinφ), where

A = |11+5| =√(11^{2}+5^{2} )=√(121+25)=√146 = 12.08

φ = arg ( 11 + 5i) = arctan(5/11) = 0.427

Then

√(11+5i) = √A (cos(φ/2)+isin(φ/2))

√A=√12.083= 3.476

φ/2=0.213

cos(φ/2)=0.977

sin(φ/2)=0.212

Hence

√(11+5i)=3.476 (0.977+i 0.212)= 3.397 + i 0.736

√(11+5i)=r(cos +isinφ)

We have that

11+5i=A(cosφ+isinφ), where

A = |11+5| =√(11

φ = arg ( 11 + 5i) = arctan(5/11) = 0.427

Then

√(11+5i) = √A (cos(φ/2)+isin(φ/2))

√A=√12.083= 3.476

φ/2=0.213

cos(φ/2)=0.977

sin(φ/2)=0.212

Hence

√(11+5i)=3.476 (0.977+i 0.212)= 3.397 + i 0.736

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