wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find all complex numbers z which satisfy the following equation
(iz)(1+2i)+(1zi)(34i)=1+7i

Open in App
Solution

Rewriting the given equation, we have
z(12i3i4)+(i2+34i17i)=0
or z=10i5(1+i)= 2i(1i)1+1 =1i

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon