Answer:
11) [tex]r=4[/tex]
12) [tex]z_3+z_1=A[/tex]
13a) [tex]\bar z_1=3-2i[/tex] and [tex]\bar z_2=-5+3i[/tex]
13b) [tex]\bar {z_1z_2}=-9-19i[/tex]
Step-by-step explanation:
11. The given equation in rectangular coordinates is [tex]x^2+y^2=16[/tex]
To convert to polar form, use the relation [tex]x^2+y^2=r^2[/tex]
This implies that:
[tex]\implies r^2=16[/tex]
[tex]r=\sqrt{16}[/tex]
[tex]r=4[/tex]
In polar coordinates, [tex]r=4[/tex] is a circle with radius 4 units.
12. From the diagram;
[tex]z_3=-3-3i[/tex]
[tex]z_2=1+i[/tex]
[tex]z_1=4+i[/tex]
[tex]A=1-2i[/tex]
We can see that two arrows are moving in anticlockwise direction to meet A which is moving in the clockwise direction;
[tex]\implies (-3-3i)+(4+i)=1-2i[/tex]
[tex]\therefore z_3+z_1=A[/tex]
13. The conjugate of the complex number: [tex]z=a+bi[/tex] is [tex]\bar z=a-bi[/tex]
Part a) The given complex numbers are:
[tex]z_1=3+2i[/tex] and [tex]z_2=-5-3i[/tex]
The conjugates of these complex numbers are:
[tex]\bar z_1=3-2i[/tex] and [tex]\bar z_2=-5+3i[/tex]
Part b) The product of [tex]z_1=3+2i[/tex] and [tex]z_2=-5-3i[/tex] is
[tex]z_1z_2=(3+2i)(-5-3i)[/tex]
[tex]z_1z_2=-15-9i-10i-6i^2[/tex]
[tex]z_1z_2=-9-19i[/tex]
The conjugate of the product is:
[tex]\bar {z_1z_2}=-9-19i[/tex]
