MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 10
        1.Find the number of solutions of Re(z^2)=0 and |z|=a
6 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
							Ans:
Hello Student,
Please find answer to your question below,

z = x + iy
z^{2} = (x + iy)^{2}
z^{2} = x^{2} + (iy)^{2} + 2ixy
z^{2} = (x^{2} - y^{2}) + 2ixy
Re(z^{2}) = x^{2} - y^{2}
Re(z^{2}) = x^{2} - y^{2} = 0
x^{2} - y^{2} = 0............(1)
|z| = a
|z|^{2} = a^{2}
x^{2} + y^{2} = a^{2}
Put y from (1), we have
x^{2} + x^{2} = a^{2}
x^{2} = \frac{a^{2}}{2}
x = \pm \frac{a}{\sqrt{2}}
y = \pm x
y = \pm \frac{a}{\sqrt{2}}
So there are four solutions
(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}), (\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}}),(-\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}) and (-\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}})
4 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 233 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details