Question 177505
*[tex \large r=\left| z \right|=\sqrt{a^2+b^2}]


*[tex \large \varphi =\arg (z)=\arctan (\frac {b}{a})]

 


So for 6 - 2j, {{{a = 6}}} and {{{b = -2}}}, so *[tex \large r=\left| z \right|=\sqrt{6^2+(-2)^2}=sqrt {40}=2sqrt{10}]


*[tex \large \varphi =\arg (z)=\arctan (\frac {-2}{6})=\arctan(-\frac{1}{3})]


Polar form notation:  *[tex \large z = r(\cos{\varphi} + i\sin{\varphi})]



*[tex \large z = 2sqrt{10}(\cos{(\arctan(-\frac{1}{3}))} + j\sin{{(\arctan(-\frac{1}{3})))} ]


Exponential notation:

*[tex \large \cos{\varphi} + i\sin{\varphi}=e^{i\varphi}], so:


*[tex \large z = re^{i\varphi}]


So for 6 - 2j, *[tex \large z = (2sqrt{10})e^{j\arctan(-\frac {1}{3})]


For your other problem:  {{{a = 0}}} and {{{b = -4}}}


Do it the same way, except that arctan(b/a) is {{{pi/2}}} because b/a is undefined.