Complex numbers into 2 different polar forms .. HELP!

Thanks a lot!

You are a life saver 

You can verify your answers easily with calculator: e^ia = cos(a) + i*sin(a)..

I'm unfamiliar with the notation used in the last one, so I can't comment on that one

On the -5i one....think about where the terminal point of this angle lies.....it's on the y axis, 5 units "down" from the origin in the "imaginary" direction {pardon the non-math terms, here}

So, the correct angle should be -pi.2, not pi/2......

Rewrite to polar form re ^ iθ, where θ is an angle in radians

$$\small{\text{$\boxed{z=a+bi \quad \to \quad re^{i\phi} }$}}\\\\ \small{\text{ $ \boxed{r=\sqrt{a^2+b^2} \qquad \phi= \left\{ \begin{array}{ll} \arctan{( \frac{b}{a} )}, & a > 0 \\ \\ \arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\ \arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\  \frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\ -\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\ \end{array} \right. } $ }}$$

I.

    $$\\\small{\text{\ $z = -3$}}\\ \small{\text{ $a=-3, \ b = 0 $}}\\ \small{\text{ $r=\sqrt{(-3)^2+0^2} = 3$}} \\ \small{\text{ $a < 0, \ b=0:\quad$ $\phi=\arctan{( \frac{b}{a} )} + \pi =\arctan{( \frac{0}{-3} )}+\pi = 0+\pi = \pi $}}\\ \small{\text{$\textcolor[rgb]{1,0,0}{z = -3 = 3e^{i\pi}}$}}$$

II.

$$\\\small{\text{\ $z = 1-i$}}\\ \small{\text{ $a=1, \ b = -1$}}\\ \small{\text{ $r=\sqrt{(1)^2+(-1)^2} = \sqrt{2}$}} \\ \small{\text{ $a > 0, \ b < 0:\quad$ $\phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-1}{1} )} = -\frac{\pi}{4} $}}\\ \small{\text{$\textcolor[rgb]{1,0,0}{z = 1-i = \sqrt{2}e^{-\frac{\pi}{4} i}}$}}$$

III.

$$\\\small{\text{\ $z = -5i$}}\\ \small{\text{ $a=0, \ b = -5$}}\\ \small{\text{ $r=\sqrt{(0)^2+(-5)^2} = 5$}} \\ \small{\text{ $a = 0, \ b < 0:\quad$ $\phi= -\frac{ \pi }{2} $}}\\ \small{\text{$\textcolor[rgb]{1,0,0}{z = -5i = 5e^{-\frac{ \pi }{2} i}}$}}$$

Thanks everyone! 

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

r = lzl  = √(a^2 + b^2) = √4^2+(-2i)^2 = √4^2+2^2 = √16+4 = √20 = 4,47

b < 0 and a > 0

θ = tan-1(b/a) + pi = tan-1(-2/4) + pi = -23,42 degrees

4,47 and -23,42 degrees

The last one is correct.

.

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

$$\small{\text{$\boxed{r=\sqrt{a^2+b^2}\qquad\phi= \left\{ \begin{array}{ll}\arctan{( \frac{b}{a} )}, & a > 0 \text{ and } b \text{ whatever!} \\ \\\arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\\arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\\frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\-\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\\end{array}\right.}$}}$$

$$\\\small{\text{\ $z = 4-2i$}}\\ \small{\text{ $a=4, \ b = -2$}}\\ \small{\text{ $r=\sqrt{(4)^2+(-2)^2} = \sqrt{20}=4.47$}} \\ \small{\text{ $a > 0 :\quad$ $\phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-2}{4} )} = -26.5650511771\ensurement{^{\circ}} $}}$$