Question

Let z_1=4-3i and z_2=2+3i. Find a complex number w, written in the form a+bi, such that wz1=z2. (Hint: First find the multiplicative inverse of z1 and then multiply both sides of the equation by that multiplicative inverse.)

A) w=2+3i/25 B) w=2-3i/25 C) w=2+3i/7 D) w=2-3i/7

Answer 1

To find the complex number w that satisfies the equation wz1 = z2, find the multiplicative inverse of z1 and multiply both sides of the equation by that inverse. The complex number w is (2-3i)/25.

Step-by-step explanation:

To find the complex number w that satisfies the equation wz1 = z2, we need to find the multiplicative inverse of z1 and multiply both sides of the equation by that inverse.

The multiplicative inverse of z1 = 4-3i is (1/z1) = 1/(4-3i). To find this inverse, we multiply the numerator and denominator by the conjugate of z1, which is 4+3i.

So, (1/z1) = (4+3i)/(4^2 - (3i)^2) = (4+3i)/(16 + 9) = (4+3i)/25.

Multiplying both sides of the equation wz1 = z2 by (1/z1), we get w = z2/z1 = (2+3i)/(4-3i) = (2+3i)*(4+3i)/(16+9) = (8+10i+9i-12)/(25) = (2-3i)/25.

Therefore, the complex number w is given by w = (2-3i)/25.