$z=\frac{1}{a+bi}$

$z=\frac{1}{r\left(\cos\theta+i\sin\theta\right)}$

$z=a+bi$

$z=r\left(\cos\theta+i\sin\theta\right)$

Spare part

Imaginary part

Real part

Illusionary part

 $-1$  

 $1$  

 $\sqrt{-1}$  

 $-\sqrt{1}$  

 $3$  

 $-3$  

 $3i$  

 $-3i$  

 $\frac{3\sqrt{3}}{2}-\frac{18}{7}i$  

 $\sqrt{3}+\frac{18}{7}i$  

 $\sqrt{3}+\frac{12}{7}i$  

 $\frac{3\sqrt{3}}{2}-\frac{12}{7}i$  

 $-12i-6i^2$  

 $-12i-6$  

 $-12i+6i^2$  

 $6-12i$  

Bring the denominator to the other side

Change the sign of the imaginary part at the denominator

Rationalise by multiplying with conjugate

 $-2-3i$  

 $2-3i$  

 $-2+3i$  

 $2+3i$  

 $a=\frac{2}{13},\ b=\frac{3}{13}$  

 $a=2,\ b=3$  

 $a=-\frac{2}{5},\ b=-\frac{3}{5}$  

 $a=-2,\ b=-3$  

2 + i

-3

2i

-1 - i

A

B

C

D

$r=\sqrt{a^2-b^2}$

$r=\sqrt{a^2-\left(bi\right)^2}$

$r=\sqrt{a^2+b^2}$

$r=\sqrt{a^2+\left(bi\right)^2}$

 $\sqrt{2}$  

 $0$  

 $2$  

 $1$  

 $\theta=-\alpha$  in the 4th quadrant

 $\theta=-\left(\pi-\alpha\right)$  in the 3rd quadrant

 $\theta=\pi-\alpha$  in the 2nd quadrant

 $\theta=\alpha$  in the 1st quadrant

 $-0.7854$  

 $-0.8$ 

 $0.7854$  

 $0.8$  

 $-2.3562$  

 $2.3562$  

 $0.7854$  

 $-0.7854$  

$z=\frac{1}{a+bi}$

$z=\frac{1}{r\left(\cos\theta+i\sin\theta\right)}$

$z=a+bi$

$z=r\left(\cos\theta+i\sin\theta\right)$

 $\sqrt{2}\left(\sin0.7854+i\cos0.7854\right)$  

 $\sqrt{2}\left(\sin\left(-2.3562\right)+i\cos\left(-2.3562\right)\right)$  

 $\sqrt{2}\left(\cos0.7854+i\sin0.7854\right)$  

 $\sqrt{2}\left(\cos\left(-2.3562\right)+i\sin\left(-2.3562\right)\right)$   

Okay jak miss

Saya masih confuse sikit2

Saya tak faham apa2 langsung miss