convert -3i into polar form.

Let {tex}-3i= 0-3i=r\left( cos\theta +isin\theta \right) {/tex}

Comparing real and imaginary parts

{tex}rcos\theta =0.......................\left( i \right) \\ rsin\theta =-3....................\left( ii \right) \\ {/tex}

Squaring and adding equations (i) and (ii)

{tex}{ r }^{ 2 }{ cos }^{ 2 }\theta +{ { r }^{ 2 } }sin^{ 2 }\theta =9\\ \Rightarrow { r }^{ 2 }=9\quad \quad \quad \quad \quad \quad \quad \left( { \because cos }^{ 2 }\theta +sin^{ 2 }\theta =1 \right) {/tex}

{tex}\Rightarrow r=3{/tex}

Substituting {tex}r=3{/tex} in equations (i) and (ii), we get 

{tex}cos\theta =0\quad and\quad sin\theta =-1\\ {/tex}

So {tex}\theta {/tex} lies in fourth quadrant 

{tex}\therefore \quad \theta =\frac { -\Pi }{ 2 } {/tex}     [ in the fourth quarant format of amplitude or principal argument is {tex}-\theta {/tex} (where {tex}\theta =\frac { \Pi }{ 2 } {/tex} ) ]

{tex}-3i=3\left( cos\left( \frac { -\Pi }{ 2 } \right) +isin\left( \frac { -\Pi }{ 2 } \right) \right) \\ \Rightarrow -3i=3\left( cos\left( \frac { \Pi }{ 2 } \right) -isin\left( \frac { \Pi }{ 2 } \right) \right) \quad \quad \left[ \because cos\left( -\theta \right) =cos\theta \quad ,sin\left( -\theta \right) =-sin\theta \right] {/tex}