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

d.) yes, because the volume of the cone is greater than the volume of the sphere

step-by-step explanation: