Question

exponential and polar form of 1+2i/1_3i​

Answer 1

Answer:

-1/2 + 1i/2 = (1/2)e^i(-45)

-1/2 + 1i/2 = 1/2(-cos(45) + isin(45))

Step-by-step explanation:

We are given

   ( 1 + 2i ) / ( 1 - 3i )

By rationalization we get

= ( 1 + 2i )( 1 + 3i ) / ( 1 - 3i )( 1 + 3i )

= ( 1 + 2i + 3i + 6i² )  / ( 1² - (3i)² )

= ( 1 - 6 + 5i )  / ( 1 + 9 )

= ( -5 + 5i ) / 10

= -1/2 + 1i/2

Now

if we compare it with

x + iy

then

x = -1/2

y = 1/2

And

we  know that

r = x² + y² = (-1/2)² + (1/2)² = ( 1 + 1 ) / 4 = 2 / 4 = 1 / 2

so

r = 1/2

We know that

tan(Ф) = y / x = (1/2) / (-1/2) = -1

tan(Ф) = -1

Taking tan inverse on both side we get

(Ф) = -45°

Since (Ф) is in the second quadrant so

x = -(1/2)cos(45)

y = (1/2)Sin(45)

So

In polar form we know that

x + iy = r(cos(Ф) + isin(Ф))

so

-1/2 + 1i/2 = 1/2(-cos(45) + isin(45))

We know that

exponential form of complex number z is given as

z =  r e^ i θ

So

-1/2 + 1i/2 = (1/2)e^i(-45)