Question

additive inverse of {3/2-i-4+i/3+5i}[6/2+i​

Answer 1

Answer:

Step-by-step explanation:

Solution :

We know that the multiplicative inverse of the complex number z is given by z−1=z¯|z|2.

(i) Let z=(5–√+3i). Then,

z¯=(5–√−3i)and|z|2=(5–√)2+32=(5+9)=14.

∴ z−1=z¯|z|2=(5–√−3i)14=(5–√14−314i).

(ii) Let z = 4 - 3i. Then,

z¯=(4−3i)¯¯¯¯¯¯¯¯¯¯¯=(4+3i)and|z|2={(4)2+(−3)2}=(16+9)=25.

∴ zz−1=z¯|z|2=(4+3i)25=(425+325i).

(iii) Let z=(3i−1)2=(9i2+1−6i)=(−9+1−6i)=(−8−6i).

∴ z¯=(−8−6i)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=(−8+6i)and|z|2={(−8)2+(−6)2}=(64+36)=100.

Hence, z−1=z¯|z|2=(−8+6i)100=(−8100+6100i)=(−225+350i)

(iv) Let z = (0-i). Then,

z¯=(0−i)¯¯¯¯¯¯¯¯¯¯=iand|z|2=02+(−1)2−−−−−−−−−−√=1.

∴ z−1=z¯|z|2=i1=i.