The complex numbers z1 and z2 are given by z1 = p +2i and z2 = 1 – 2i , where p is an integer

p = ±21

Step-by-step explanation:

Given  z1 = p +2i and z2 = 1 – 2i, If |z1/z2| = 13;

|p+2i/1-21| = 13

To get p from the expression above, we need to rationalize the complex function first.

p+2i/1-21 = p+2i/1-2i * 1+2i/1+2i

= p+2pi+2i+4i²/1-4i²

Since i² = -1;

= p+2pi+2i-4/1+4

= p-4+2i(p+1)/5

= p-4/5 + 2(p+1)/5 i

Then we will take the modulus of the resulting expression and equate to the value of 13 to get p

|p+2i/1-21| = √(p-4/5)²+ (2p+2/5)² = 13

(p-4/5)²+ (2p+2/5)² = 13²

(p-4)²+(2p+2)² = 13²*5²

p²-8p+16+4p²+8p+4 = 4225

5p²+20 = 4225

5p² = 4205

p² = 841

p = ±√841

p = ±21

The possible values of p are 21 and -21

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

dont think so

step-by-step explanation: