Given z1=6(cos45degrees+isin45degrees) and z2=2(cos90degrees+isin90degrees), find z1/z2 when 0<theta<360degrees

Answer 1

To find z1/z2, we need to divide z1 by z2.

Firstly, let's write z1 and z2 in exponential form:

z1 = 6(cos(45°) + isin(45°)) = 6(cos(45°) + i(sin(45°)) = 6cis(45°) = 6e^(i*45°)

z2 = 2(cos(90°) + isin(90°)) = 2(cos(90°) + i(sin(90°)) = 2cis(90°) = 2e^(i*90°)

Now, to divide z1 by z2:

z1/z2 = (6e^(i*45°))/(2e^(i*90°))
= (6/2)e^(i*(45°-90°))
= 3e^(-i*45°)
= 3(cos(-45°) + isin(-45°))
= 3(cos(315°) + isin(315°))

Therefore, z1/z2 = 3(cos(315°) + isin(315°)) in exponential form.