Question

find the multiplication inverse of the following complex number = (2-3i)^2​

Answer 1

Step-by-step explanation:

Given :(2-3i) ^2

• 4+9i^2-12i (i^2=-1)
• 4+9(-1)-12i
• 4-9-12i
• -5-12i

rationalising on both sides with -5+12i

• let multiplicative inverse of z=1/z
• (1/(-5-12i))(-5+12i/-5+12i)
• -5+12i/(-5)^2-(12i)^2 (i^2=-1)
• -5+12i/25-144(-1)
• -5+12i/25+144
• -5+12i/169