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

(i) Find the real values of x and y for which z1=9y2410ix is complex conjugate of each other.
(ii) Find the value of x4x3+x2+3x5 if x=2+3i

Open in App
Solution

1. ¯z1=z
9y24+10ix=9y2410ix
20ix=0
x=0
x=0 & yϵR, this holds true
2. x4x3+x2+3x5=P (Let)
x=2+3i
x2=3i
(x2)2=9i2
x24x+4=9
x24x+13=0
(Image)
P=(x24x+13)(x2+3x)+(36x5)
Remainder=Answer=36x5
=36(2+3i)5=77108i

996592_1029635_ans_83db8c337f794883964f647e48388d11.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Properties of Conjugate of a Complex Number
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon