Question

find the poler and exponential from of 1+2i​

Answer 1

-1/2 + 1i/2 $&#x3d$ (1/2)e^i(-45)

-1/2 + 1i/2$&#x3d$1/2(-cos(45) + isin(45))

Step-by-step explanation:

We are given

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

By rationalization we get:

$&#x3d$ ( 1 + 2i )( 1 + 3i ) / ( 1 - 3i )( 1 + 3i )

$&#x3d$( 1 + 2i + 3i + 6i² ) / ( 1² - (3i)² )

$&#x3d$( 1 - 6 + 5i ) / ( 1 + 9 )

$&#x3d$( -5 + 5i ) / 10

$&#x3d$-1/2 + 1i/2

Now,

if we compare it with

x + iy

then

x$&#x3d$-1/2

y$&#x3d$1/2

And

we know that,

r$&#x3d$x² + y² = (-1/2)² + (1/2)²$&#x3d$( 1 + 1 ) / 4$&#x3d$2 / 4$&#x3d$1 / 2

so

r$&#x3d$1/2

we know that,

tan(Ф)$&#x3d$y / x$&#x3d$(1/2) / (-1/2)$&#x3d$-1

tan(Ф)$&#x3d$-1

Taking tan inverse on both side we get

(Ф)$&#x3d$-45°

Since (Ф) is in the second quadrant

so,

x$&#x3d$-(1/2)cos(45)

y$&#x3d$(1/2)Sin(45)

so,

In polar form we know that

x + iy$&#x3d$r(cos(Ф) + isin(Ф))

so,

-1/2 + 1i/2$&#x3d$1/2(-cos(45) + isin(45))

We know that,

exponential form of complex number z is given as

z$&#x3d$r e^ i θ

So,

-1/2 + 1i/2$&#x3d$(1/2)e^i(-45)