Math, asked by harshitpanchal818, 5 months ago

find the principal argument and general argument for
1 +  \sqrt{3}i

Answers

Answered by Hulk12343
0

Answer:

3x + y

Step-by-step explanation:

it is the answer... m

Answered by Anonymous
0

Required Answer :-

Given,

Given,z=5+5i=a+ib

Given,z=5+5i=a+ibPrinciple argument,

Given,z=5+5i=a+ibPrinciple argument,θ=tan

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 (

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( a

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 (

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 5

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan −1

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan −1 (1)

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan −1 (1)∴θ=

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan −1 (1)∴θ= 4

Given,z=5+5i=a+ibPrinciple argument,θ=tan −1 ( ab )=tan −1 ( 55 )=tan −1 (1)∴θ= 4π

Have a great day ahead mate ❤️

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